Explanation:
Okay, so I need to solve this recurrence relation problem. Let me try to understand what's given and what's being asked.

The problem says that for $ n \geq 1 $, the function $ f $ satisfies the recurrence relation:
$$
f(n + 1) = (-1)^{n + 1} n - 2f(n)
$$
and also that $ f(1) = f(1986) $. We are to compute the sum $ f(1) + f(2) + f(3) + \dots + f(1985) $.

First, let me note that the function is defined for integers $ n \geq 1 $, and it's periodic in some sense because $ f(1) = f(1986) $. That suggests that maybe the sequence is periodic with period 1985? Or perhaps there's some other periodicity? Not sure yet. But maybe I can find a general formula for $ f(n) $, or find a way to express the sum in terms of itself?

Let me start by writing down the recurrence relation again:
$$
f(n + 1) + 2f(n) = (-1)^{n + 1} n
$$
This looks like a linear nonhomogeneous recurrence relation. Maybe I can solve it using standard techniques for such equations. Let me recall that for linear recurrences of the form $ f(n + 1) + a f(n) = g(n) $, we can use methods like finding the homogeneous solution and a particular solution.

First, let me write the recurrence in standard form. Let me rearrange the equation:
$$
f(n + 1) = -2f(n) + (-1)^{n + 1} n
$$
So the homogeneous part is $ f(n + 1) + 2f(n) = 0 $, and the nonhomogeneous part is $ (-1)^{n + 1} n $.

To solve this, I need to find the general solution to the homogeneous equation and then find a particular solution to the nonhomogeneous equation.

First, solving the homogeneous equation:
$$
f(n + 1) + 2f(n) = 0
$$
This is a linear homogeneous recurrence relation with constant coefficients. The characteristic equation would be $ r + 2 = 0 $, so $ r = -2 $. Therefore, the general solution to the homogeneous equation is:
$$
f_h(n) = C(-2)^{n - 1}
$$
Wait, let me check that. If the characteristic equation is $ r = -2 $, then the solution is $ f_h(n) = C(-2)^{n} $, but depending on how we index it. Let me think. Let me suppose that the homogeneous recurrence is $ f(n + 1) = -2f(n) $. Then the solution is $ f(n) = f(1)(-2)^{n - 1} $. So yes, the general solution is $ C(-2)^{n} $, but maybe with a different constant. Anyway, the homogeneous solution is exponential with base -2.

Now, for the particular solution. The nonhomogeneous term is $ (-1)^{n + 1} n $. Let me note that $ (-1)^{n + 1} = -(-1)^n $, so the nonhomogeneous term is $ -(-1)^n n $. So the nonhomogeneous term is of the form $ (-1)^n \times n $, which is a polynomial multiplied by an exponential (in this case, the exponential is $ (-1)^n $, which is like $ r^n $ with $ r = -1 $).

Therefore, to find a particular solution, I can use the method of undetermined coefficients. However, since the nonhomogeneous term is $ (-1)^n n $, and since $ r = -1 $ is not a root of the homogeneous equation (the homogeneous equation has root $ r = -2 $), so we can try a particular solution of the form $ (An + B)(-1)^n $.

Wait, let me check. The nonhomogeneous term is $ (-1)^n n $, so if we assume a particular solution of the form $ (An + B)(-1)^n $, then we can plug it into the recurrence and solve for A and B.

Let me denote the particular solution as $ f_p(n) = (An + B)(-1)^n $.

Let me compute $ f_p(n + 1) $:
$$
f_p(n + 1) = [A(n + 1) + B](-1)^{n + 1} = -[A(n + 1) + B](-1)^n
$$
Now, according to the recurrence relation:
$$
f_p(n + 1) + 2f_p(n) = (-1)^{n + 1} n
$$
Substituting the expressions:
Left-hand side (LHS):
$$
- [A(n + 1) + B](-1)^n + 2(An + B)(-1)^n
$$
Factor out $ (-1)^n $:
$$
[ -A(n + 1) - B + 2An + 2B ] (-1)^n
$$
Simplify inside the brackets:
- $ -A(n + 1) - B + 2An + 2B = -An - A - B + 2An + 2B = ( -An + 2An ) + ( -A - B + 2B ) = An + (-A + B) $

Therefore, the LHS becomes:
$$
[An + (-A + B)] (-1)^n
$$
This is supposed to equal the right-hand side (RHS), which is $ (-1)^{n + 1} n = -(-1)^n n $.

Therefore, equating the two sides:
$$
[An + (-A + B)] (-1)^n = -n (-1)^n
$$
Divide both sides by $ (-1)^n $ (which is non-zero):
$$
An + (-A + B) = -n
$$
Now, equate coefficients for like terms. Let me write the left side as $ An + ( -A + B ) $, and the right side is $ -n + 0 $.

Therefore, equating coefficients:

For the coefficient of n:
$ A = -1 $

For the constant term:
$ -A + B = 0 $

Substituting A = -1 into the second equation:
$ -(-1) + B = 0 \implies 1 + B = 0 \implies B = -1 $

Therefore, the particular solution is:
$$
f_p(n) = (-n - 1)(-1)^n = (-1)^n (-n - 1) = (-1)^{n + 1}(n + 1)
$$
Wait, let me check that again. If $ f_p(n) = (An + B)(-1)^n $, with A = -1 and B = -1, then:

$ f_p(n) = (-n - 1)(-1)^n = (-1)^n (-n - 1) = (-1)^{n + 1}(n + 1) $. Yes, that's correct.

Alternatively, maybe I can write it as $ f_p(n) = (-1)^{n + 1}(n + 1) $.

Therefore, the general solution to the recurrence is the homogeneous solution plus the particular solution:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

Wait, let me check. The homogeneous solution was $ C(-2)^{n} $, and the particular solution is $ (-1)^{n + 1}(n + 1) $. So combining them:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

But let me check if this works. Let me test for a small n. Let me take n=1. Wait, but maybe I need to check the recurrence.

Alternatively, maybe I can write the general solution as:

$$
f(n) = C(-2)^{n} + (-1)^{n + 1}(n + 1)
$$

But let me check if this satisfies the recurrence. Let me compute f(n + 1) + 2f(n):

First, compute f(n + 1):

$$
f(n + 1) = C(-2)^{n + 1} + (-1)^{(n + 1) + 1}((n + 1) + 1) = C(-2)^{n + 1} + (-1)^{n + 2}(n + 2)
$$

Then compute 2f(n):

$$
2f(n) = 2C(-2)^n + 2(-1)^{n + 1}(n + 1)
$$

Adding them together:

f(n + 1) + 2f(n) = C(-2)^{n + 1} + (-1)^{n + 2}(n + 2) + 2C(-2)^n + 2(-1)^{n + 1}(n + 1)

Let me simplify term by term:

First term: C(-2)^{n + 1} = -2C(-2)^n

Second term: (-1)^{n + 2}(n + 2) = (-1)^n (-1)^2 (n + 2) = (-1)^n (n + 2)

Third term: 2C(-2)^n

Fourth term: 2(-1)^{n + 1}(n + 1) = -2(-1)^n (n + 1)

So combining:

-2C(-2)^n + 2C(-2)^n + (-1)^n (n + 2) - 2(-1)^n (n + 1)

The first two terms cancel out. Then:

(-1)^n [ (n + 2) - 2(n + 1) ] = (-1)^n [n + 2 - 2n - 2] = (-1)^n (-n) = -n(-1)^n = (-1)^{n + 1} n

Which matches the RHS of the original recurrence. Great! So the general solution is correct.

Therefore, the general solution is:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

Now, we need to determine the constant C. However, we have the condition that $ f(1) = f(1986) $. So we can use this to find C.

But first, let me write expressions for f(1) and f(1986) using the general solution.

First, compute f(1):

$$
f(1) = C(-2)^1 + (-1)^{1 + 1}(1 + 1) = -2C + (-1)^2 \times 2 = -2C + 1 \times 2 = -2C + 2
$$

Now compute f(1986):

$$
f(1986) = C(-2)^{1986} + (-1)^{1986 + 1}(1986 + 1) = C(-2)^{1986} + (-1)^{1987} \times 1987
$$

Note that $ (-1)^{1987} = -1 $, since 1987 is odd. Therefore:

$$
f(1986) = C(-2)^{1986} - 1987
$$

But since $ f(1) = f(1986) $, we have:

$$
-2C + 2 = C(-2)^{1986} - 1987
$$

Let me rearrange this equation to solve for C:

Bring all terms to one side:

$$
-2C + 2 - C(-2)^{1986} + 1987 = 0
$$

Wait, actually, let me write:

$$
-2C + 2 = C(-2)^{1986} - 1987
$$

Bring all terms to the left:

$$
-2C + 2 - C(-2)^{1986} + 1987 = 0
$$

Combine constants:

2 + 1987 = 1989

So:

$$
-2C - C(-2)^{1986} + 1989 = 0
$$

Factor out C:

$$
C[ -2 - (-2)^{1986} ] + 1989 = 0
$$

Therefore:

$$
C = \frac{ -1989 }{ -2 - (-2)^{1986} } = \frac{1989}{2 + (-2)^{1986}}
$$

Now, let me compute the denominator. Let's note that $ (-2)^{1986} $. Since 1986 is even, this is equal to $ 2^{1986} $. Therefore:

Denominator: $ 2 + 2^{1986} = 2(1 + 2^{1985}) $

Therefore:

$$
C = \frac{1989}{2(1 + 2^{1985})}
$$

Hmm, that seems complicated. But maybe we don't need to compute C explicitly? Because we need to compute the sum $ S = f(1) + f(2) + \dots + f(1985) $. Maybe there's a way to express this sum in terms of C, or maybe the terms involving C will cancel out?

Alternatively, perhaps there's a smarter way to compute the sum without finding C explicitly. Let me think.

Alternatively, maybe I can use the recurrence relation to find a relation for the sum S.

Let me denote $ S = \sum_{n=1}^{1985} f(n) $. Our goal is to compute S.

Given that the recurrence is $ f(n + 1) = (-1)^{n + 1} n - 2f(n) $, perhaps I can sum both sides over n from 1 to 1985.

Let me try that.

Summing both sides from n = 1 to n = 1985:

Left-hand side (LHS): $ \sum_{n=1}^{1985} f(n + 1) = \sum_{n=2}^{1986} f(n) = \sum_{n=1}^{1986} f(n) - f(1) $

Right-hand side (RHS): $ \sum_{n=1}^{1985} [ (-1)^{n + 1} n - 2f(n) ] = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2 \sum_{n=1}^{1985} f(n) $

Therefore, equating LHS and RHS:

$$
\sum_{n=1}^{1986} f(n) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2S
$$

But note that $ \sum_{n=1}^{1986} f(n) = S + f(1986) $, since S is the sum from 1 to 1985. Therefore:

Left-hand side becomes:

$ S + f(1986) - f(1) $

Therefore, the equation is:

$$
S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2S
$$

Now, rearranging terms:

Bring all S terms to the left and others to the right:

$ S + 2S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n $

Which simplifies to:

$ 3S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n $

But we know from the problem statement that $ f(1) = f(1986) $, so $ f(1986) - f(1) = 0 $. Therefore:

$ 3S = \sum_{n=1}^{1985} (-1)^{n + 1} n $

Therefore, $ S = \frac{1}{3} \sum_{n=1}^{1985} (-1)^{n + 1} n $

So now, the problem reduces to computing this alternating sum. Let me compute that sum.

Let me denote $ T = \sum_{n=1}^{1985} (-1)^{n + 1} n $

Note that $ (-1)^{n + 1} = -(-1)^n $, so:

$ T = - \sum_{n=1}^{1985} (-1)^n n $

But maybe it's easier to compute directly. Let me think about the sum of an alternating series of integers.

Let me note that the sum $ \sum_{n=1}^m (-1)^{n + 1} n $ can be evaluated depending on whether m is even or odd.

Let me recall that for such sums:

If m is even, say m = 2k:

$ \sum_{n=1}^{2k} (-1)^{n + 1} n = \sum_{i=1}^k [ (2i - 1) - 2i ] = \sum_{i=1}^k (-1) = -k $

If m is odd, say m = 2k + 1:

$ \sum_{n=1}^{2k + 1} (-1)^{n + 1} n = \sum_{n=1}^{2k} (-1)^{n + 1} n + (-1)^{2k + 2} (2k + 1) = -k + (1)(2k + 1) = -k + 2k + 1 = k + 1 $

Therefore, in general:

If m is even, the sum is -m/2

If m is odd, the sum is (m + 1)/2

Wait, let me check with small m:

For m = 1: sum is 1. According to formula, (1 + 1)/2 = 1. Correct.

For m = 2: 1 - 2 = -1. According to formula, -2/2 = -1. Correct.

For m = 3: 1 - 2 + 3 = 2. According to formula, (3 + 1)/2 = 2. Correct.

For m = 4: 1 - 2 + 3 - 4 = -2. According to formula, -4/2 = -2. Correct. So yes, the formula holds.

Therefore, in our case, m = 1985, which is odd. Therefore, the sum T is (1985 + 1)/2 = 1986 / 2 = 993.

Therefore, T = 993. Therefore, S = (1/3) * 993 = 331.

Wait, so the answer is 331? Let me check again.

Wait, let me confirm:

Given that m = 1985 is odd, so T = (m + 1)/2 = (1985 + 1)/2 = 1986 / 2 = 993. Then S = T / 3 = 993 / 3 = 331. So yes, 331.

But let me check again the steps to make sure I didn't make a mistake.

First, we had the recurrence relation, found the general solution, then tried to compute the sum S by summing the recurrence. Then, through that process, we found that S = (1/3) * T, where T is the alternating sum. Then, since m = 1985 is odd, T = (1985 + 1)/2 = 993, so S = 993 / 3 = 331. Seems straightforward.

But let me check if there was a mistake in the earlier steps. Let me go back.

We had:

Summing the recurrence from n=1 to 1985:

LHS: sum_{n=1}^{1985} f(n+1) = sum_{n=2}^{1986} f(n) = sum_{n=1}^{1986} f(n) - f(1)

RHS: sum_{n=1}^{1985} [ (-1)^{n+1} n - 2f(n) ] = sum (-1)^{n+1} n - 2 sum f(n)

Then equate:

sum_{n=1}^{1986} f(n) - f(1) = sum (-1)^{n+1} n - 2S

But sum_{n=1}^{1986} f(n) = S + f(1986). Therefore:

S + f(1986) - f(1) = sum (-1)^{n+1} n - 2S

Then moving terms:

S + 2S + f(1986) - f(1) = sum (-1)^{n+1} n

Which is 3S + (f(1986) - f(1)) = sum...

But since f(1) = f(1986), that term is zero, so 3S = sum... which is T. Therefore, S = T / 3. Then T is computed as 993, so S = 331. Seems correct.

But let me check with a small case. Let me take a smaller number instead of 1985, say m = 1. Wait, but m = 1 would mean sum from n=1 to 1, but the original problem has f(1) = f(1986). Let me try with m = 1, but maybe not. Let me try with m = 3, so that the period is 3? Wait, maybe not. Let me think of a small case where I can compute manually.

Suppose instead of 1985, let me take N = 1, but then f(1) = f(2). Let me try N = 1. Wait, maybe not. Let me take N = 2. Suppose the problem was f(1) = f(3), compute sum f(1) + f(2). Let me try with N = 2.

Wait, let me try with N = 2. Let me define f(1) = f(3). Let me compute the sum S = f(1) + f(2).

First, let me use the recurrence:

For n = 1: f(2) = (-1)^{2} * 1 - 2f(1) = 1*1 - 2f(1) = 1 - 2f(1)

For n = 2: f(3) = (-1)^{3} * 2 - 2f(2) = -2 - 2f(2)

But since f(3) = f(1), so:

f(1) = -2 - 2f(2)

But from the first equation, f(2) = 1 - 2f(1). Substitute into the second equation:

f(1) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

Therefore, f(1) - 4f(1) = -4 => -3f(1) = -4 => f(1) = 4/3

Then f(2) = 1 - 2*(4/3) = 1 - 8/3 = -5/3

Therefore, the sum S = f(1) + f(2) = 4/3 - 5/3 = -1/3

Now, according to our formula, T would be sum_{n=1}^{N} (-1)^{n + 1} n where N = 2? Wait, no. Wait in the original problem, the sum S was up to N = 1985, which was the number of terms. Wait in this case, if we have N = 2, then the sum S would be up to n = 2? Wait, but in the original problem, the sum was up to 1985, which was one less than 1986. So in this case, if we have f(1) = f(3), then the sum is up to n = 2. So in this case, m = 2, which is even. Then according to our formula, T = sum_{n=1}^2 (-1)^{n + 1} n = 1 - 2 = -1. Then S = T / 3 = -1 / 3, which matches the manual calculation. So that works.

Another test case: Let me take N = 3, so that f(1) = f(4). Compute sum S = f(1) + f(2) + f(3).

First, compute using recurrence:

For n=1: f(2) = (-1)^{2}*1 - 2f(1) = 1 - 2f(1)

For n=2: f(3) = (-1)^{3}*2 - 2f(2) = -2 - 2f(2)

For n=3: f(4) = (-1)^{4}*3 - 2f(3) = 3 - 2f(3)

But f(4) = f(1), so:

f(1) = 3 - 2f(3)

From above, f(3) = -2 - 2f(2), and f(2) = 1 - 2f(1). Substitute:

f(3) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

Then f(1) = 3 - 2*(-4 + 4f(1)) = 3 + 8 - 8f(1) = 11 - 8f(1)

Therefore, f(1) + 8f(1) = 11 => 9f(1) = 11 => f(1) = 11/9

Then f(2) = 1 - 2*(11/9) = 1 - 22/9 = -13/9

f(3) = -4 + 4*(11/9) = -4 + 44/9 = (-36 + 44)/9 = 8/9

Sum S = 11/9 -13/9 + 8/9 = (11 -13 + 8)/9 = 6/9 = 2/3

Now, according to our formula, T = sum_{n=1}^3 (-1)^{n + 1} n = 1 - 2 + 3 = 2. Then S = T / 3 = 2 / 3. Which matches. Perfect!

Therefore, our formula seems to hold. Therefore, in the original problem, since m = 1985 is odd, T = (1985 + 1)/2 = 993, so S = 993 / 3 = 331. Therefore, the answer is 331.

But let me check once more if there was a miscalculation in the T value. For m = 1985, which is odd, the formula says (m + 1)/2 = 1986 / 2 = 993. Yes. Then 993 divided by 3 is indeed 331. So that seems correct.

Alternatively, maybe I can check with another approach. Let me think.

Alternatively, since we have the general solution for f(n):

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

Then, the sum S = sum_{n=1}^{1985} f(n) = C sum_{n=1}^{1985} (-2)^n + sum_{n=1}^{1985} (-1)^{n + 1}(n + 1)

We can compute both sums separately.

First, compute sum_{n=1}^{1985} (-2)^n. This is a geometric series.

Sum_{n=1}^m r^n = r(1 - r^m)/(1 - r). Here, r = -2, m = 1985.

Therefore:

Sum = (-2)(1 - (-2)^{1985}) / (1 - (-2)) = (-2)(1 - (-2)^{1985}) / 3

Similarly, the second sum is sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{n=1}^{1985} (-1)^{n + 1}(n + 1)

Let me make substitution k = n + 1. When n = 1, k = 2; when n = 1985, k = 1986. Therefore:

sum_{k=2}^{1986} (-1)^{k} k = sum_{k=1}^{1986} (-1)^k k - (-1)^1 * 1 = [sum_{k=1}^{1986} (-1)^k k] + 1

But note that sum_{k=1}^{m} (-1)^k k is equal to -T where T was our previous sum. Wait, earlier we had T = sum_{n=1}^m (-1)^{n + 1} n = -sum_{n=1}^m (-1)^n n. Therefore, sum_{k=1}^{1986} (-1)^k k = -T', where T' is the sum for m = 1986. But since m = 1986 is even, T' = -1986 / 2 = -993. Therefore, sum_{k=1}^{1986} (-1)^k k = -T' = 993. Therefore, the second sum is 993 + 1 = 994? Wait, let me check again.

Wait, original sum is sum_{k=2}^{1986} (-1)^k k = sum_{k=1}^{1986} (-1)^k k - (-1)^1 * 1 = [sum_{k=1}^{1986} (-1)^k k] - (-1) * 1 = [sum_{k=1}^{1986} (-1)^k k] + 1

Earlier, we said that sum_{k=1}^{m} (-1)^k k = -T where T was sum_{n=1}^m (-1)^{n + 1} n. For m = 1986, which is even, T = -m / 2 = -1986 / 2 = -993. Therefore, sum_{k=1}^{1986} (-1)^k k = -T = 993. Therefore, the second sum is 993 + 1 = 994?

Wait, but earlier we had that T = sum_{n=1}^{1985} (-1)^{n + 1} n = 993. But here, the second sum is sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{k=2}^{1986} (-1)^k k = 994? Wait, but according to our previous approach, the sum S was equal to T / 3 = 993 / 3 = 331. But if we compute via this method, we have:

S = C * [ (-2)(1 - (-2)^{1985}) / 3 ] + 994

But we also have from the general solution that f(1) = -2C + 2, and f(1986) = C(-2)^{1986} - 1987, and since f(1) = f(1986), we had found C = 1989 / [2 + 2^{1986}]. But this seems complicated. However, if we use the previous result that S = 331, then according to this approach, we can check if the two expressions for S are consistent.

But maybe this is getting too complicated. However, since we already verified with small cases and the logic seems sound, and the answer 331 seems consistent, I think that is correct.

Alternatively, let me check with the general solution. Let me compute S = sum_{n=1}^{1985} f(n) = C * sum_{n=1}^{1985} (-2)^n + sum_{n=1}^{1985} (-1)^{n + 1}(n + 1)

We already found that the second sum is 994? Wait, earlier I thought it was 994, but let me recast:

sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{k=2}^{1986} (-1)^{k} k = [sum_{k=1}^{1986} (-1)^k k] - (-1)^1 * 1 = [sum_{k=1}^{1986} (-1)^k k] + 1

As before, sum_{k=1}^{1986} (-1)^k k = -T', where T' is the sum for m = 1986, which is even, so T' = -m / 2 = -1986 / 2 = -993. Therefore, sum_{k=1}^{1986} (-1)^k k = -T' = 993. Therefore, the second sum is 993 + 1 = 994. Therefore, S = C * [ (-2)(1 - (-2)^{1985}) / 3 ] + 994.

But we also have from the previous approach that S = 331. Therefore:

331 = C * [ (-2)(1 - (-2)^{1985}) / 3 ] + 994

Therefore, solving for C:

C * [ (-2)(1 - (-2)^{1985}) / 3 ] = 331 - 994 = -663

Therefore,

C = (-663) / [ (-2)(1 - (-2)^{1985}) / 3 ] = (-663) * [ 3 / (-2)(1 - (-2)^{1985}) ] = (663 * 3) / [ 2(1 - (-2)^{1985}) ]

But this seems complicated, but maybe we can check if this is consistent with our previous expression for C.

Earlier, we had:

C = 1989 / [ 2 + 2^{1986} ]

But note that 2^{1986} = (-2)^{1986} since 1986 is even. Therefore, denominator is 2 + (-2)^{1986} = 2 + 2^{1986}

So C = 1989 / [ 2 + 2^{1986} ]

But let me compute the expression from the second approach:

C = (663 * 3) / [ 2(1 - (-2)^{1985}) ] = 1989 / [ 2(1 - (-2)^{1985}) ]

Therefore, equate the two expressions for C:

1989 / [ 2 + 2^{1986} ] = 1989 / [ 2(1 - (-2)^{1985}) ]

Therefore, denominators must be equal:

2 + 2^{1986} = 2(1 - (-2)^{1985})

Let me check:

Left side: 2 + 2^{1986}

Right side: 2(1 - (-2)^{1985}) = 2 - 2*(-2)^{1985}

Note that (-2)^{1985} = -2^{1985} because 1985 is odd. Therefore:

Right side: 2 - 2*(-(-2^{1985})) = 2 - 2*(-(-2^{1985}))? Wait:

Wait, (-2)^{1985} = (-1)^{1985} * 2^{1985} = -2^{1985}

Therefore, right side: 2 - 2*(-(-2^{1985}))? Wait no:

Wait, right side is 2 - 2*(-2)^{1985} = 2 - 2*(-2^{1985}) = 2 + 2*2^{1985} = 2 + 2^{1986}

Which is equal to the left side. Therefore, yes, the denominators are equal. Therefore, the expressions for C are consistent. Therefore, this confirms that our calculation of S = 331 is correct.

Therefore, the answer is 331.

**Final Answer**
\boxed{331}

To solve the recurrence relation:

$$
f(n + 1) = (-1)^{n + 1} n - 2f(n), \quad \text{for } n \geq 1,
$$
with the condition $ f(1) = f(1986) $, we are asked to compute the sum:

$$
S = f(1) + f(2) + \cdots + f(1985).
$$

---

### Step 1: General Solution of the Recurrence

We begin by solving the recurrence relation. The recurrence is linear and nonhomogeneous:

$$
f(n + 1) + 2f(n) = (-1)^{n + 1} n.
$$

This is a first-order linear recurrence. The general solution is the sum of the homogeneous and particular solutions.

#### Homogeneous Solution

The homogeneous recurrence is:

$$
f(n + 1) + 2f(n) = 0.
$$

This has the characteristic equation $ r + 2 = 0 $, so the solution is:

$$
f_h(n) = C(-2)^n,
$$
where $ C $ is a constant.

#### Particular Solution

The nonhomogeneous term is $ (-1)^{n + 1} n $. We try a particular solution of the form:

$$
f_p(n) = (An + B)(-1)^n.
$$

Substituting into the recurrence and solving for $ A $ and $ B $, we find:

$$
f_p(n) = (-1)^{n + 1}(n + 1).
$$

#### General Solution

Combining both parts, the general solution is:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1).
$$

---

### Step 2: Use the Periodicity Condition

We are given that $ f(1) = f(1986) $. Using the general solution:

- $ f(1) = C(-2)^1 + (-1)^{2}(1 + 1) = -2C + 2 $
- $ f(1986) = C(-2)^{1986} + (-1)^{1987}(1986 + 1) = C(2^{1986}) - 1987 $

Setting $ f(1) = f(1986) $:

$$
-2C + 2 = C(2^{1986}) - 1987.
$$

Solving for $ C $:

$$
C = \frac{1989}{2 + 2^{1986}}.
$$

---

### Step 3: Compute the Sum $ S = \sum_{n=1}^{1985} f(n) $

We use the general solution:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1).
$$

So the sum becomes:

$$
S = \sum_{n=1}^{1985} f(n) = C \sum_{n=1}^{1985} (-2)^n + \sum_{n=1}^{1985} (-1)^{n + 1}(n + 1).
$$

Let’s compute each part.

#### Part 1: Geometric Sum

$$
\sum_{n=1}^{1985} (-2)^n = \frac{-2(1 - (-2)^{1985})}{1 - (-2)} = \frac{-2(1 - (-2)^{1985})}{3}.
$$

#### Part 2: Alternating Sum

We compute:

$$
\sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = \sum_{k=2}^{1986} (-1)^k k = \sum_{k=1}^{1986} (-1)^k k - (-1)^1 \cdot 1 = \sum_{k=1}^{1986} (-1)^k k + 1.
$$

Now, for even $ m = 1986 $, the sum $ \sum_{k=1}^{m} (-1)^k k = -\frac{m}{2} = -993 $, so:

$$
\sum_{k=1}^{1986} (-1)^k k = -993 \Rightarrow \sum_{k=2}^{1986} (-1)^k k = -993 + 1 = -992.
$$

Thus:

$$
\sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = -992.
$$

---

### Step 4: Combine the Results

Now, we have:

$$
S = C \cdot \frac{-2(1 - (-2)^{1985})}{3} - 992.
$$

But instead of computing this directly, we can use a more elegant approach by summing the recurrence directly.

---

### Step 5: Sum the Recurrence

We sum both sides of the recurrence from $ n = 1 $ to $ n = 1985 $:

$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{n=1}^{1985} \left[ (-1)^{n + 1} n - 2f(n) \right].
$$

Left-hand side:

$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{n=2}^{1986} f(n) = \sum_{n=1}^{1986} f(n) - f(1).
$$

Right-hand side:

$$
\sum_{n=1}^{1985} (-1)^{n + 1} n - 2 \sum_{n=1}^{1985} f(n).
$$

Let $ S = \sum_{n=1}^{1985} f(n) $, and note that $ \sum_{n=1}^{1986} f(n) = S + f(1986) $. Then:

$$
S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2S.
$$

Rearranging:

$$
3S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n.
$$

But since $ f(1) = f(1986) $, the difference is zero, so:

$$
3S = \sum_{n=1}^{1985} (-1)^{n + 1} n.
$$

---

### Step 6: Evaluate the Alternating Sum

We now compute:

$$
\sum_{n=1}^{1985} (-1)^{n + 1} n.
$$

This is a well-known alternating sum of the first $ m $ natural numbers. For odd $ m $, the sum is:

$$
\sum_{n=1}^{m} (-1)^{n + 1} n = \frac{m + 1}{2}.
$$

Since $ m = 1985 $ is odd:

$$
\sum_{n=1}^{1985} (-1)^{n + 1} n = \frac{1985 + 1}{2} = 993.
$$

Thus:

$$
3S = 993 \Rightarrow S = \frac{993}{3} = 331.
$$

---

### Final Answer

$$
\boxed{331}
$$

[Assistant]
Explanation:
Okay, so I need to solve this recurrence relation problem. Let me try to understand what's given and what's being asked.

The problem says that for $ n \geq 1 $, the function $ f $ satisfies the recurrence relation:
$$
f(n + 1) = (-1)^{n + 1} n - 2f(n)
$$
and also that $ f(1) = f(1986) $. We are to compute the sum $ f(1) + f(2) + f(3) + \dots + f(1985) $.

First, let me note that the function is defined for integers $ n \geq 1 $, and it's periodic in some sense because $ f(1) = f(1986) $. That suggests that maybe the sequence is periodic with period 1985? Or perhaps there's some other periodicity? Not sure yet. But maybe I can find a general formula for $ f(n) $, or find a way to express the sum in terms of itself?

Let me start by writing down the recurrence relation again:
$$
f(n + 1) + 2f(n) = (-1)^{n + 1} n
$$
This looks like a linear nonhomogeneous recurrence relation. Maybe I can solve it using standard techniques for such equations. Let me recall that for linear recurrences of the form $ f(n + 1) + a f(n) = g(n) $, we can use methods like finding the homogeneous solution and a particular solution.

First, let me write the recurrence in standard form. Let me rearrange the equation:
$$
f(n + 1) = -2f(n) + (-1)^{n + 1} n
$$
So the homogeneous part is $ f(n + 1) + 2f(n) = 0 $, and the nonhomogeneous part is $ (-1)^{n + 1} n $.

To solve this, I need to find the general solution to the homogeneous equation and then find a particular solution to the nonhomogeneous equation.

First, solving the homogeneous equation:
$$
f(n + 1) + 2f(n) = 0
$$
This is a linear homogeneous recurrence relation with constant coefficients. The characteristic equation would be $ r + 2 = 0 $, so $ r = -2 $. Therefore, the general solution to the homogeneous equation is:
$$
f_h(n) = C(-2)^{n - 1}
$$
Wait, let me check that. If the characteristic equation is $ r = -2 $, then the solution is $ f_h(n) = C(-2)^{n} $, but depending on how we index it. Let me think. Let me suppose that the homogeneous recurrence is $ f(n + 1) = -2f(n) $. Then the solution is $ f(n) = f(1)(-2)^{n - 1} $. So yes, the general solution is $ C(-2)^{n} $, but maybe with a different constant. Anyway, the homogeneous solution is exponential with base -2.

Now, for the particular solution. The nonhomogeneous term is $ (-1)^{n + 1} n $. Let me note that $ (-1)^{n + 1} = -(-1)^n $, so the nonhomogeneous term is $ -(-1)^n n $. So the nonhomogeneous term is of the form $ (-1)^n \times n $, which is a polynomial multiplied by an exponential (in this case, the exponential is $ (-1)^n $, which is like $ r^n $ with $ r = -1 $).

Therefore, to find a particular solution, I can use the method of undetermined coefficients. However, since the nonhomogeneous term is $ (-1)^n n $, and since $ r = -1 $ is not a root of the homogeneous equation (the homogeneous equation has root $ r = -2 $), so we can try a particular solution of the form $ (An + B)(-1)^n $.

Wait, let me check. The nonhomogeneous term is $ (-1)^n n $, so if we assume a particular solution of the form $ (An + B)(-1)^n $, then we can plug it into the recurrence and solve for A and B.

Let me denote the particular solution as $ f_p(n) = (An + B)(-1)^n $.

Let me compute $ f_p(n + 1) $:
$$
f_p(n + 1) = [A(n + 1) + B](-1)^{n + 1} = -[A(n + 1) + B](-1)^n
$$
Now, according to the recurrence relation:
$$
f_p(n + 1) + 2f_p(n) = (-1)^{n + 1} n
$$
Substituting the expressions:
Left-hand side (LHS):
$$
- [A(n + 1) + B](-1)^n + 2(An + B)(-1)^n
$$
Factor out $ (-1)^n $:
$$
[ -A(n + 1) - B + 2An + 2B ] (-1)^n
$$
Simplify inside the brackets:
- $ -A(n + 1) - B + 2An + 2B = -An - A - B + 2An + 2B = ( -An + 2An ) + ( -A - B + 2B ) = An + (-A + B) $

Therefore, the LHS becomes:
$$
[An + (-A + B)] (-1)^n
$$
This is supposed to equal the right-hand side (RHS), which is $ (-1)^{n + 1} n = -(-1)^n n $.

Therefore, equating the two sides:
$$
[An + (-A + B)] (-1)^n = -n (-1)^n
$$
Divide both sides by $ (-1)^n $ (which is non-zero):
$$
An + (-A + B) = -n
$$
Now, equate coefficients for like terms. Let me write the left side as $ An + ( -A + B ) $, and the right side is $ -n + 0 $.

Therefore, equating coefficients:

For the coefficient of n:
$ A = -1 $

For the constant term:
$ -A + B = 0 $

Substituting A = -1 into the second equation:
$ -(-1) + B = 0 \implies 1 + B = 0 \implies B = -1 $

Therefore, the particular solution is:
$$
f_p(n) = (-n - 1)(-1)^n = (-1)^n (-n - 1) = (-1)^{n + 1}(n + 1)
$$
Wait, let me check that again. If $ f_p(n) = (An + B)(-1)^n $, with A = -1 and B = -1, then:

$ f_p(n) = (-n - 1)(-1)^n = (-1)^n (-n - 1) = (-1)^{n + 1}(n + 1) $. Yes, that's correct.

Alternatively, maybe I can write it as $ f_p(n) = (-1)^{n + 1}(n + 1) $.

Therefore, the general solution to the recurrence is the homogeneous solution plus the particular solution:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

Wait, let me check. The homogeneous solution was $ C(-2)^{n} $, and the particular solution is $ (-1)^{n + 1}(n + 1) $. So combining them:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

But let me check if this works. Let me test for a small n. Let me take n=1. Wait, but maybe I need to check the recurrence.

Alternatively, maybe I can write the general solution as:

$$
f(n) = C(-2)^{n} + (-1)^{n + 1}(n + 1)
$$

But let me check if this satisfies the recurrence. Let me compute f(n + 1) + 2f(n):

First, compute f(n + 1):

$$
f(n + 1) = C(-2)^{n + 1} + (-1)^{(n + 1) + 1}((n + 1) + 1) = C(-2)^{n + 1} + (-1)^{n + 2}(n + 2)
$$

Then compute 2f(n):

$$
2f(n) = 2C(-2)^n + 2(-1)^{n + 1}(n + 1)
$$

Adding them together:

f(n + 1) + 2f(n) = C(-2)^{n + 1} + (-1)^{n + 2}(n + 2) + 2C(-2)^n + 2(-1)^{n + 1}(n + 1)

Let me simplify term by term:

First term: C(-2)^{n + 1} = -2C(-2)^n

Second term: (-1)^{n + 2}(n + 2) = (-1)^n (-1)^2 (n + 2) = (-1)^n (n + 2)

Third term: 2C(-2)^n

Fourth term: 2(-1)^{n + 1}(n + 1) = -2(-1)^n (n + 1)

So combining:

-2C(-2)^n + 2C(-2)^n + (-1)^n (n + 2) - 2(-1)^n (n + 1)

The first two terms cancel out. Then:

(-1)^n [ (n + 2) - 2(n + 1) ] = (-1)^n [n + 2 - 2n - 2] = (-1)^n (-n) = -n(-1)^n = (-1)^{n + 1} n

Which matches the RHS of the original recurrence. Great! So the general solution is correct.

Therefore, the general solution is:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

Now, we need to determine the constant C. However, we have the condition that $ f(1) = f(1986) $. So we can use this to find C.

But first, let me write expressions for f(1) and f(1986) using the general solution.

First, compute f(1):

$$
f(1) = C(-2)^1 + (-1)^{1 + 1}(1 + 1) = -2C + (-1)^2 \times 2 = -2C + 1 \times 2 = -2C + 2
$$

Now compute f(1986):

$$
f(1986) = C(-2)^{1986} + (-1)^{1986 + 1}(1986 + 1) = C(-2)^{1986} + (-1)^{1987} \times 1987
$$

Note that $ (-1)^{1987} = -1 $, since 1987 is odd. Therefore:

$$
f(1986) = C(-2)^{1986} - 1987
$$

But since $ f(1) = f(1986) $, we have:

$$
-2C + 2 = C(-2)^{1986} - 1987
$$

Let me rearrange this equation to solve for C:

Bring all terms to one side:

$$
-2C + 2 - C(-2)^{1986} + 1987 = 0
$$

Wait, actually, let me write:

$$
-2C + 2 = C(-2)^{1986} - 1987
$$

Bring all terms to the left:

$$
-2C + 2 - C(-2)^{1986} + 1987 = 0
$$

Combine constants:

2 + 1987 = 1989

So:

$$
-2C - C(-2)^{1986} + 1989 = 0
$$

Factor out C:

$$
C[ -2 - (-2)^{1986} ] + 1989 = 0
$$

Therefore:

$$
C = \frac{ -1989 }{ -2 - (-2)^{1986} } = \frac{1989}{2 + (-2)^{1986}}
$$

Now, let me compute the denominator. Let's note that $ (-2)^{1986} $. Since 1986 is even, this is equal to $ 2^{1986} $. Therefore:

Denominator: $ 2 + 2^{1986} = 2(1 + 2^{1985}) $

Therefore:

$$
C = \frac{1989}{2(1 + 2^{1985})}
$$

Hmm, that seems complicated. But maybe we don't need to compute C explicitly? Because we need to compute the sum $ S = f(1) + f(2) + \dots + f(1985) $. Maybe there's a way to express this sum in terms of C, or maybe the terms involving C will cancel out?

Alternatively, perhaps there's a smarter way to compute the sum without finding C explicitly. Let me think.

Alternatively, maybe I can use the recurrence relation to find a relation for the sum S.

Let me denote $ S = \sum_{n=1}^{1985} f(n) $. Our goal is to compute S.

Given that the recurrence is $ f(n + 1) = (-1)^{n + 1} n - 2f(n) $, perhaps I can sum both sides over n from 1 to 1985.

Let me try that.

Summing both sides from n = 1 to n = 1985:

Left-hand side (LHS): $ \sum_{n=1}^{1985} f(n + 1) = \sum_{n=2}^{1986} f(n) = \sum_{n=1}^{1986} f(n) - f(1) $

Right-hand side (RHS): $ \sum_{n=1}^{1985} [ (-1)^{n + 1} n - 2f(n) ] = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2 \sum_{n=1}^{1985} f(n) $

Therefore, equating LHS and RHS:

$$
\sum_{n=1}^{1986} f(n) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2S
$$

But note that $ \sum_{n=1}^{1986} f(n) = S + f(1986) $, since S is the sum from 1 to 1985. Therefore:

Left-hand side becomes:

$ S + f(1986) - f(1) $

Therefore, the equation is:

$$
S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2S
$$

Now, rearranging terms:

Bring all S terms to the left and others to the right:

$ S + 2S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n $

Which simplifies to:

$ 3S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n $

But we know from the problem statement that $ f(1) = f(1986) $, so $ f(1986) - f(1) = 0 $. Therefore:

$ 3S = \sum_{n=1}^{1985} (-1)^{n + 1} n $

Therefore, $ S = \frac{1}{3} \sum_{n=1}^{1985} (-1)^{n + 1} n $

So now, the problem reduces to computing this alternating sum. Let me compute that sum.

Let me denote $ T = \sum_{n=1}^{1985} (-1)^{n + 1} n $

Note that $ (-1)^{n + 1} = -(-1)^n $, so:

$ T = - \sum_{n=1}^{1985} (-1)^n n $

But maybe it's easier to compute directly. Let me think about the sum of an alternating series of integers.

Let me note that the sum $ \sum_{n=1}^m (-1)^{n + 1} n $ can be evaluated depending on whether m is even or odd.

Let me recall that for such sums:

If m is even, say m = 2k:

$ \sum_{n=1}^{2k} (-1)^{n + 1} n = \sum_{i=1}^k [ (2i - 1) - 2i ] = \sum_{i=1}^k (-1) = -k $

If m is odd, say m = 2k + 1:

$ \sum_{n=1}^{2k + 1} (-1)^{n + 1} n = \sum_{n=1}^{2k} (-1)^{n + 1} n + (-1)^{2k + 2} (2k + 1) = -k + (1)(2k + 1) = -k + 2k + 1 = k + 1 $

Therefore, in general:

If m is even, the sum is -m/2

If m is odd, the sum is (m + 1)/2

Wait, let me check with small m:

For m = 1: sum is 1. According to formula, (1 + 1)/2 = 1. Correct.

For m = 2: 1 - 2 = -1. According to formula, -2/2 = -1. Correct.

For m = 3: 1 - 2 + 3 = 2. According to formula, (3 + 1)/2 = 2. Correct.

For m = 4: 1 - 2 + 3 - 4 = -2. According to formula, -4/2 = -2. Correct. So yes, the formula holds.

Therefore, in our case, m = 1985, which is odd. Therefore, the sum T is (1985 + 1)/2 = 1986 / 2 = 993.

Therefore, T = 993. Therefore, S = (1/3) * 993 = 331.

Wait, so the answer is 331? Let me check again.

Wait, let me confirm:

Given that m = 1985 is odd, so T = (m + 1)/2 = (1985 + 1)/2 = 1986 / 2 = 993. Then S = T / 3 = 993 / 3 = 331. So yes, 331.

But let me check again the steps to make sure I didn't make a mistake.

First, we had the recurrence relation, found the general solution, then tried to compute the sum S by summing the recurrence. Then, through that process, we found that S = (1/3) * T, where T is the alternating sum. Then, since m = 1985 is odd, T = (1985 + 1)/2 = 993, so S = 993 / 3 = 331. Seems straightforward.

But let me check if there was a mistake in the earlier steps. Let me go back.

We had:

Summing the recurrence from n=1 to 1985:

LHS: sum_{n=1}^{1985} f(n+1) = sum_{n=2}^{1986} f(n) = sum_{n=1}^{1986} f(n) - f(1)

RHS: sum_{n=1}^{1985} [ (-1)^{n+1} n - 2f(n) ] = sum (-1)^{n+1} n - 2 sum f(n)

Then equate:

sum_{n=1}^{1986} f(n) - f(1) = sum (-1)^{n+1} n - 2S

But sum_{n=1}^{1986} f(n) = S + f(1986). Therefore:

S + f(1986) - f(1) = sum (-1)^{n+1} n - 2S

Then moving terms:

S + 2S + f(1986) - f(1) = sum (-1)^{n+1} n

Which is 3S + (f(1986) - f(1)) = sum...

But since f(1) = f(1986), that term is zero, so 3S = sum... which is T. Therefore, S = T / 3. Then T is computed as 993, so S = 331. Seems correct.

But let me check with a small case. Let me take a smaller number instead of 1985, say m = 1. Wait, but m = 1 would mean sum from n=1 to 1, but the original problem has f(1) = f(1986). Let me try with m = 1, but maybe not. Let me try with m = 3, so that the period is 3? Wait, maybe not. Let me think of a small case where I can compute manually.

Suppose instead of 1985, let me take N = 1, but then f(1) = f(2). Let me try N = 1. Wait, maybe not. Let me take N = 2. Suppose the problem was f(1) = f(3), compute sum f(1) + f(2). Let me try with N = 2.

Wait, let me try with N = 2. Let me define f(1) = f(3). Let me compute the sum S = f(1) + f(2).

First, let me use the recurrence:

For n = 1: f(2) = (-1)^{2} * 1 - 2f(1) = 1*1 - 2f(1) = 1 - 2f(1)

For n = 2: f(3) = (-1)^{3} * 2 - 2f(2) = -2 - 2f(2)

But since f(3) = f(1), so:

f(1) = -2 - 2f(2)

But from the first equation, f(2) = 1 - 2f(1). Substitute into the second equation:

f(1) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

Therefore, f(1) - 4f(1) = -4 => -3f(1) = -4 => f(1) = 4/3

Then f(2) = 1 - 2*(4/3) = 1 - 8/3 = -5/3

Therefore, the sum S = f(1) + f(2) = 4/3 - 5/3 = -1/3

Now, according to our formula, T would be sum_{n=1}^{N} (-1)^{n + 1} n where N = 2? Wait, no. Wait in the original problem, the sum S was up to N = 1985, which was the number of terms. Wait in this case, if we have N = 2, then the sum S would be up to n = 2? Wait, but in the original problem, the sum was up to 1985, which was one less than 1986. So in this case, if we have f(1) = f(3), then the sum is up to n = 2. So in this case, m = 2, which is even. Then according to our formula, T = sum_{n=1}^2 (-1)^{n + 1} n = 1 - 2 = -1. Then S = T / 3 = -1 / 3, which matches the manual calculation. So that works.

Another test case: Let me take N = 3, so that f(1) = f(4). Compute sum S = f(1) + f(2) + f(3).

First, compute using recurrence:

For n=1: f(2) = (-1)^{2}*1 - 2f(1) = 1 - 2f(1)

For n=2: f(3) = (-1)^{3}*2 - 2f(2) = -2 - 2f(2)

For n=3: f(4) = (-1)^{4}*3 - 2f(3) = 3 - 2f(3)

But f(4) = f(1), so:

f(1) = 3 - 2f(3)

From above, f(3) = -2 - 2f(2), and f(2) = 1 - 2f(1). Substitute:

f(3) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

Then f(1) = 3 - 2*(-4 + 4f(1)) = 3 + 8 - 8f(1) = 11 - 8f(1)

Therefore, f(1) + 8f(1) = 11 => 9f(1) = 11 => f(1) = 11/9

Then f(2) = 1 - 2*(11/9) = 1 - 22/9 = -13/9

f(3) = -4 + 4*(11/9) = -4 + 44/9 = (-36 + 44)/9 = 8/9

Sum S = 11/9 -13/9 + 8/9 = (11 -13 + 8)/9 = 6/9 = 2/3

Now, according to our formula, T = sum_{n=1}^3 (-1)^{n + 1} n = 1 - 2 + 3 = 2. Then S = T / 3 = 2 / 3. Which matches. Perfect!

Therefore, our formula seems to hold. Therefore, in the original problem, since m = 1985 is odd, T = (1985 + 1)/2 = 993, so S = 993 / 3 = 331. Therefore, the answer is 331.

But let me check once more if there was a miscalculation in the T value. For m = 1985, which is odd, the formula says (m + 1)/2 = 1986 / 2 = 993. Yes. Then 993 divided by 3 is indeed 331. So that seems correct.

Alternatively, maybe I can check with another approach. Let me think.

Alternatively, since we have the general solution for f(n):

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1)
$$

Then, the sum S = sum_{n=1}^{1985} f(n) = C sum_{n=1}^{1985} (-2)^n + sum_{n=1}^{1985} (-1)^{n + 1}(n + 1)

We can compute both sums separately.

First, compute sum_{n=1}^{1985} (-2)^n. This is a geometric series.

Sum_{n=1}^m r^n = r(1 - r^m)/(1 - r). Here, r = -2, m = 1985.

Therefore:

Sum = (-2)(1 - (-2)^{1985}) / (1 - (-2)) = (-2)(1 - (-2)^{1985}) / 3

Similarly, the second sum is sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{n=1}^{1985} (-1)^{n + 1}(n + 1)

Let me make substitution k = n + 1. When n = 1, k = 2; when n = 1985, k = 1986. Therefore:

sum_{k=2}^{1986} (-1)^{k} k = sum_{k=1}^{1986} (-1)^k k - (-1)^1 * 1 = [sum_{k=1}^{1986} (-1)^k k] + 1

But note that sum_{k=1}^{m} (-1)^k k is equal to -T where T was our previous sum. Wait, earlier we had T = sum_{n=1}^m (-1)^{n + 1} n = -sum_{n=1}^m (-1)^n n. Therefore, sum_{k=1}^{1986} (-1)^k k = -T', where T' is the sum for m = 1986. But since m = 1986 is even, T' = -1986 / 2 = -993. Therefore, sum_{k=1}^{1986} (-1)^k k = -T' = 993. Therefore, the second sum is 993 + 1 = 994? Wait, let me check again.

Wait, original sum is sum_{k=2}^{1986} (-1)^k k = sum_{k=1}^{1986} (-1)^k k - (-1)^1 * 1 = [sum_{k=1}^{1986} (-1)^k k] - (-1) * 1 = [sum_{k=1}^{1986} (-1)^k k] + 1

Earlier, we said that sum_{k=1}^{m} (-1)^k k = -T where T was sum_{n=1}^m (-1)^{n + 1} n. For m = 1986, which is even, T = -m / 2 = -1986 / 2 = -993. Therefore, sum_{k=1}^{1986} (-1)^k k = -T = 993. Therefore, the second sum is 993 + 1 = 994?

Wait, but earlier we had that T = sum_{n=1}^{1985} (-1)^{n + 1} n = 993. But here, the second sum is sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{k=2}^{1986} (-1)^k k = 994? Wait, but according to our previous approach, the sum S was equal to T / 3 = 993 / 3 = 331. But if we compute via this method, we have:

S = C * [ (-2)(1 - (-2)^{1985}) / 3 ] + 994

But we also have from the general solution that f(1) = -2C + 2, and f(1986) = C(-2)^{1986} - 1987, and since f(1) = f(1986), we had found C = 1989 / [2 + 2^{1986}]. But this seems complicated. However, if we use the previous result that S = 331, then according to this approach, we can check if the two expressions for S are consistent.

But maybe this is getting too complicated. However, since we already verified with small cases and the logic seems sound, and the answer 331 seems consistent, I think that is correct.

Alternatively, let me check with the general solution. Let me compute S = sum_{n=1}^{1985} f(n) = C * sum_{n=1}^{1985} (-2)^n + sum_{n=1}^{1985} (-1)^{n + 1}(n + 1)

We already found that the second sum is 994? Wait, earlier I thought it was 994, but let me recast:

sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = sum_{k=2}^{1986} (-1)^{k} k = [sum_{k=1}^{1986} (-1)^k k] - (-1)^1 * 1 = [sum_{k=1}^{1986} (-1)^k k] + 1

As before, sum_{k=1}^{1986} (-1)^k k = -T', where T' is the sum for m = 1986, which is even, so T' = -m / 2 = -1986 / 2 = -993. Therefore, sum_{k=1}^{1986} (-1)^k k = -T' = 993. Therefore, the second sum is 993 + 1 = 994. Therefore, S = C * [ (-2)(1 - (-2)^{1985}) / 3 ] + 994.

But we also have from the previous approach that S = 331. Therefore:

331 = C * [ (-2)(1 - (-2)^{1985}) / 3 ] + 994

Therefore, solving for C:

C * [ (-2)(1 - (-2)^{1985}) / 3 ] = 331 - 994 = -663

Therefore,

C = (-663) / [ (-2)(1 - (-2)^{1985}) / 3 ] = (-663) * [ 3 / (-2)(1 - (-2)^{1985}) ] = (663 * 3) / [ 2(1 - (-2)^{1985}) ]

But this seems complicated, but maybe we can check if this is consistent with our previous expression for C.

Earlier, we had:

C = 1989 / [ 2 + 2^{1986} ]

But note that 2^{1986} = (-2)^{1986} since 1986 is even. Therefore, denominator is 2 + (-2)^{1986} = 2 + 2^{1986}

So C = 1989 / [ 2 + 2^{1986} ]

But let me compute the expression from the second approach:

C = (663 * 3) / [ 2(1 - (-2)^{1985}) ] = 1989 / [ 2(1 - (-2)^{1985}) ]

Therefore, equate the two expressions for C:

1989 / [ 2 + 2^{1986} ] = 1989 / [ 2(1 - (-2)^{1985}) ]

Therefore, denominators must be equal:

2 + 2^{1986} = 2(1 - (-2)^{1985})

Let me check:

Left side: 2 + 2^{1986}

Right side: 2(1 - (-2)^{1985}) = 2 - 2*(-2)^{1985}

Note that (-2)^{1985} = -2^{1985} because 1985 is odd. Therefore:

Right side: 2 - 2*(-(-2^{1985})) = 2 - 2*(-(-2^{1985}))? Wait:

Wait, (-2)^{1985} = (-1)^{1985} * 2^{1985} = -2^{1985}

Therefore, right side: 2 - 2*(-(-2^{1985}))? Wait no:

Wait, right side is 2 - 2*(-2)^{1985} = 2 - 2*(-2^{1985}) = 2 + 2*2^{1985} = 2 + 2^{1986}

Which is equal to the left side. Therefore, yes, the denominators are equal. Therefore, the expressions for C are consistent. Therefore, this confirms that our calculation of S = 331 is correct.

Therefore, the answer is 331.

**Final Answer**
\boxed{331}

To solve the recurrence relation:

$$
f(n + 1) = (-1)^{n + 1} n - 2f(n), \quad \text{for } n \geq 1,
$$
with the condition $ f(1) = f(1986) $, we are asked to compute the sum:

$$
S = f(1) + f(2) + \cdots + f(1985).
$$

---

### Step 1: General Solution of the Recurrence

We begin by solving the recurrence relation. The recurrence is linear and nonhomogeneous:

$$
f(n + 1) + 2f(n) = (-1)^{n + 1} n.
$$

This is a first-order linear recurrence. The general solution is the sum of the homogeneous and particular solutions.

#### Homogeneous Solution

The homogeneous recurrence is:

$$
f(n + 1) + 2f(n) = 0.
$$

This has the characteristic equation $ r + 2 = 0 $, so the solution is:

$$
f_h(n) = C(-2)^n,
$$
where $ C $ is a constant.

#### Particular Solution

The nonhomogeneous term is $ (-1)^{n + 1} n $. We try a particular solution of the form:

$$
f_p(n) = (An + B)(-1)^n.
$$

Substituting into the recurrence and solving for $ A $ and $ B $, we find:

$$
f_p(n) = (-1)^{n + 1}(n + 1).
$$

#### General Solution

Combining both parts, the general solution is:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1).
$$

---

### Step 2: Use the Periodicity Condition

We are given that $ f(1) = f(1986) $. Using the general solution:

- $ f(1) = C(-2)^1 + (-1)^{2}(1 + 1) = -2C + 2 $
- $ f(1986) = C(-2)^{1986} + (-1)^{1987}(1986 + 1) = C(2^{1986}) - 1987 $

Setting $ f(1) = f(1986) $:

$$
-2C + 2 = C(2^{1986}) - 1987.
$$

Solving for $ C $:

$$
C = \frac{1989}{2 + 2^{1986}}.
$$

---

### Step 3: Compute the Sum $ S = \sum_{n=1}^{1985} f(n) $

We use the general solution:

$$
f(n) = C(-2)^n + (-1)^{n + 1}(n + 1).
$$

So the sum becomes:

$$
S = \sum_{n=1}^{1985} f(n) = C \sum_{n=1}^{1985} (-2)^n + \sum_{n=1}^{1985} (-1)^{n + 1}(n + 1).
$$

Let’s compute each part.

#### Part 1: Geometric Sum

$$
\sum_{n=1}^{1985} (-2)^n = \frac{-2(1 - (-2)^{1985})}{1 - (-2)} = \frac{-2(1 - (-2)^{1985})}{3}.
$$

#### Part 2: Alternating Sum

We compute:

$$
\sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = \sum_{k=2}^{1986} (-1)^k k = \sum_{k=1}^{1986} (-1)^k k - (-1)^1 \cdot 1 = \sum_{k=1}^{1986} (-1)^k k + 1.
$$

Now, for even $ m = 1986 $, the sum $ \sum_{k=1}^{m} (-1)^k k = -\frac{m}{2} = -993 $, so:

$$
\sum_{k=1}^{1986} (-1)^k k = -993 \Rightarrow \sum_{k=2}^{1986} (-1)^k k = -993 + 1 = -992.
$$

Thus:

$$
\sum_{n=1}^{1985} (-1)^{n + 1}(n + 1) = -992.
$$

---

### Step 4: Combine the Results

Now, we have:

$$
S = C \cdot \frac{-2(1 - (-2)^{1985})}{3} - 992.
$$

But instead of computing this directly, we can use a more elegant approach by summing the recurrence directly.

---

### Step 5: Sum the Recurrence

We sum both sides of the recurrence from $ n = 1 $ to $ n = 1985 $:

$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{n=1}^{1985} \left[ (-1)^{n + 1} n - 2f(n) \right].
$$

Left-hand side:

$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{n=2}^{1986} f(n) = \sum_{n=1}^{1986} f(n) - f(1).
$$

Right-hand side:

$$
\sum_{n=1}^{1985} (-1)^{n + 1} n - 2 \sum_{n=1}^{1985} f(n).
$$

Let $ S = \sum_{n=1}^{1985} f(n) $, and note that $ \sum_{n=1}^{1986} f(n) = S + f(1986) $. Then:

$$
S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n - 2S.
$$

Rearranging:

$$
3S + f(1986) - f(1) = \sum_{n=1}^{1985} (-1)^{n + 1} n.
$$

But since $ f(1) = f(1986) $, the difference is zero, so:

$$
3S = \sum_{n=1}^{1985} (-1)^{n + 1} n.
$$

---

### Step 6: Evaluate the Alternating Sum

We now compute:

$$
\sum_{n=1}^{1985} (-1)^{n + 1} n.
$$

This is a well-known alternating sum of the first $ m $ natural numbers. For odd $ m $, the sum is:

$$
\sum_{n=1}^{m} (-1)^{n + 1} n = \frac{m + 1}{2}.
$$

Since $ m = 1985 $ is odd:

$$
\sum_{n=1}^{1985} (-1)^{n + 1} n = \frac{1985 + 1}{2} = 993.
$$

Thus:

$$
3S = 993 \Rightarrow S = \frac{993}{3} = 331.
$$

---

### Final Answer

$$
\boxed{331}
$$

Solution:
We can list the equations
\begin{align*}
f(2) &= 1 - 2f(1), \\
f(3) &= -2 - 2f(2), \\
f(4) &= 3 - 2f(3), \\
f(5) &= -4 - 2f(4), \\
&\dots, \\
f(1985) &= -1984 - 2f(1984), \\
f(1986) &= 1985 - 2f(1985).
\end{align*}Adding these equations, we get
\[f(2) + f(3) + \dots + f(1986) = (1 - 2 + 3 - 4 + \dots + 1983 - 1984 + 1985) - 2f(1) - 2f(2) - \dots - 2f(1985).\]To find $1 - 2 + 3 - 4 + \dots + 1983 - 1984 + 1985,$ we can pair the terms
\begin{align*}
1 - 2 + 3 - 4 + \dots + 1983 - 1984 + 1985 &= (1 - 2) + (3 - 4) + \dots + (1983 - 1984) + 1985 \\
&= (-1) + (-1) + \dots + (-1) + 1985 \\
&= -\frac{1984}{2} + 1985 \\
&= 993.
\end{align*}Hence,
\[f(2) + f(3) + \dots + f(1986) = 993 - 2f(1) - 2f(2) - \dots - 2f(1985).\]Then
\[2f(1) + 3f(2) + 3f(3) + \dots + 3f(1985) + f(1986) = 993.\]Since $f(1986) = f(1),$
\[3f(1) + 3f(2) + 3f(3) + \dots + 3f(1985) = 993.\]Therefore, $f(1) + f(2) + f(3) + \dots + f(1985) = \boxed{331}.$


Explanation:
Okay, so I need to find the area of an isosceles right triangle where the altitude to the hypotenuse is given as 4√2. Hmm, let me think. First, let me recall some properties of isosceles right triangles.

An isosceles right triangle has two legs that are equal in length and a hypotenuse. The angles are 45°, 45°, and 90°. The area of a triangle is generally (base × height)/2. But here, they mention the altitude to the hypotenuse. So maybe I need to relate the altitude to the hypotenuse with the area?

Wait, let me remember. In any triangle, the area can be calculated using any side as the base, with the corresponding height. So if I take the hypotenuse as the base, then the altitude to the hypotenuse would be the height. Therefore, the area is (hypotenuse × altitude)/2. But I also know that the area can be calculated using the legs. Since it's an isosceles right triangle, the legs are equal. Let me denote the legs as 'a'. Then the hypotenuse would be a√2, right? Because in a right triangle, hypotenuse squared is equal to the sum of the squares of the legs. So, hypotenuse = √(a² + a²) = √(2a²) = a√2.

So, the area can be calculated in two ways: one using the legs and one using the hypotenuse and the altitude. Let me write that down.

First, area using legs: (a × a)/2 = a²/2.

Second, area using hypotenuse and altitude: (hypotenuse × altitude)/2 = (a√2 × 4√2)/2. Let me compute that.

Multiplying a√2 by 4√2: a * 4 * √2 * √2 = 4a * (√2 * √2) = 4a * 2 = 8a. Then divide by 2: 8a / 2 = 4a. So the area is 4a.

But we also have the area as a²/2. Therefore, setting them equal: a²/2 = 4a. Let me solve for a. Multiply both sides by 2: a² = 8a. Then subtract 8a: a² - 8a = 0. Factor: a(a - 8) = 0. So, a = 0 or a = 8. Since a is a length, it can't be zero, so a = 8.

Therefore, the legs are 8 each. Then the area is a²/2 = 8² / 2 = 64 / 2 = 32. So the area is 32? Wait, but let me check with the other formula. If hypotenuse is a√2 = 8√2, and altitude is 4√2, then area is (8√2 × 4√2)/2. Let me compute that again. 8√2 × 4√2 = 32 × (√2 × √2) = 32 × 2 = 64. Then divide by 2: 64 / 2 = 32. Yep, same result. So that seems consistent.

But let me think again if there's another way to approach this problem, just to verify. Maybe using properties of the altitude in an isosceles right triangle.

In an isosceles right triangle, the altitude to the hypotenuse also acts as a median and an angle bisector. Wait, is that true? In an isosceles triangle, the altitude from the right angle to the hypotenuse divides the triangle into two smaller, congruent triangles. Also, in this case, since it's a right triangle, the altitude to the hypotenuse can be calculated using some formula?

Alternatively, maybe I can recall that in a right triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. Let me check that.

Yes, in any right triangle, the altitude h to the hypotenuse is given by h = (leg1 × leg2)/hypotenuse.

So in this case, since it's an isosceles right triangle, leg1 = leg2 = a. Hypotenuse is a√2. Therefore, h = (a × a)/ (a√2) ) = a² / (a√2) ) = a / √2.

But the problem says that the altitude is 4√2. Therefore, a / √2 = 4√2. Let me solve for a. Multiply both sides by √2: a = 4√2 × √2 = 4 × 2 = 8. So again, a = 8. Then area is (a²)/2 = 64 / 2 = 32. Same answer.

So that seems to confirm it. Therefore, the area is 32.

But let me think again if there's another way someone might approach this problem, maybe leading to a different answer? Let me check.

Alternatively, suppose I didn't remember the formula for the altitude in a right triangle. Let me try to derive it.

Let me denote the triangle ABC, with right angle at C, and legs AC and BC equal. Let me call the hypotenuse AB. The altitude from C to AB is CD, where D is the foot of the altitude. Then, in triangle ABC, area is (AC × BC)/2 = (a²)/2. Also, area is (AB × CD)/2. Therefore, (AB × CD)/2 = (a²)/2. Therefore, AB × CD = a². Since AB = a√2, then CD = a² / AB = a² / (a√2) = a / √2. Which is the same as before. So CD = a / √2. Given CD = 4√2, so a / √2 = 4√2 => a = 4√2 × √2 = 4 × 2 = 8. So again, same result.

Therefore, area is 32.

Alternatively, maybe using coordinate geometry. Let me try that. Let me place the triangle on coordinate axes. Let me put point C at (0, 0), point A at (a, 0), and point B at (0, a). Then hypotenuse AB goes from (a, 0) to (0, a). The equation of AB is x + y = a. The altitude from C(0,0) to AB. Wait, but in this case, the altitude from C to AB is the same as the line from C to D, where D is the foot of the altitude.

But since ABC is a right triangle, the altitude from the right angle to hypotenuse is CD. Wait, but in coordinate terms, how do we find the length of CD?

Alternatively, since we know coordinates of A, B, and C, we can compute the equation of AB and then find the distance from C to AB.

Equation of AB: as above, x + y = a. The distance from point (0, 0) to the line x + y - a = 0 is |0 + 0 - a| / √(1² + 1²) ) = | -a | / √2 = a / √2. Which is the same as before. Therefore, the altitude is a / √2 = 4√2. Therefore, solving for a gives a = 8. Then area is (a²)/2 = 32.

Same answer again. So seems like no matter how I approach it, the area is 32.

But let me check once more. Maybe I made a mistake in assuming the altitude formula? Let me think.

Alternatively, suppose I didn't remember any formulas. Let me try to find the altitude.

In triangle ABC, right-angled at C, with AC = BC = a. Let me find the length of CD, where D is the foot of the altitude from C to AB.

In a right triangle, the altitude to hypotenuse divides the triangle into two smaller similar triangles. So triangle ABC ~ triangle ACD ~ triangle CBD.

Therefore, ratios of sides can be used.

In particular, AC / AB = AD / AC. Because triangle ABC ~ triangle ACD.

So AC / AB = AD / AC. Therefore, AC² = AB × AD.

Similarly, since AC = a, AB = a√2, then a² = a√2 × AD => AD = a² / (a√2) ) = a / √2.

Similarly, BD = BC² / AB = a² / (a√2) ) = a / √2. Therefore, AD = BD, which makes sense because the triangle is isosceles, so the altitude splits the hypotenuse into two equal parts. Therefore, AD = BD = a / √2. Therefore, AB = AD + DB = 2a / √2 = a√2, which matches the earlier result.

But how does this help me? Well, if I can find CD, which is the altitude. In triangle ACD, which is a right triangle, with legs AD and CD, and hypotenuse AC.

So, AC² = AD² + CD².

Therefore, a² = (a / √2)² + CD².

Calculating (a / √2)² = a² / 2. Therefore, a² = a² / 2 + CD² => CD² = a² - a² / 2 = a² / 2 => CD = a / √2. Which is the same as before.

So again, CD = a / √2 = 4√2 => a = 8. Then area is 32.

So all methods lead to the same answer. Therefore, I think the area is indeed 32.

But just to make sure, let me compute the hypotenuse length. If legs are 8, hypotenuse is 8√2. Then altitude is 4√2. Let me check if the area via hypotenuse and altitude is (8√2 × 4√2)/2.

Calculating numerator: 8√2 × 4√2 = 32 × (√2 × √2) = 32 × 2 = 64. Then divide by 2: 64 / 2 = 32. Which is the same as area via legs: (8 × 8)/2 = 64 / 2 = 32.

Therefore, all checks out.

Alternatively, maybe I can think of the altitude in terms of the triangle's properties. In an isosceles right triangle, the altitude to hypotenuse is half the length of the hypotenuse? Wait, no. Wait, if hypotenuse is a√2, then altitude is a / √2. Let me see if that's half the hypotenuse.

Half the hypotenuse would be (a√2)/2 = a / √2. Wait, that's exactly the altitude! Wait, so in this case, the altitude is equal to half the hypotenuse?

Wait, so altitude h = (a√2)/2. But we also have h = a / √2. Let me check if these are equal.

(a√2)/2 vs a / √2. Multiply numerator and denominator of a / √2 by √2: (a√2)/2. Yes, they are equal. Therefore, in an isosceles right triangle, the altitude to hypotenuse is equal to half the hypotenuse.

Therefore, h = (hypotenuse)/2. Therefore, if h = 4√2, then hypotenuse is 8√2. Then, since hypotenuse is a√2, then a√2 = 8√2 => a = 8. Then area is (a²)/2 = 64 / 2 = 32.

So that's another way to think about it. Since the altitude is half the hypotenuse, then hypotenuse is 8√2, so legs are 8 each, area is 32.

Therefore, regardless of the approach, the area is 32.

I think that's thorough enough. I can't see any mistakes in my reasoning. All methods converge to the same answer. So I think the area of the triangle is 32.

**Final Answer**
The area of the triangle is \boxed{32}.

To find the area of the isosceles right triangle, we are given that the **altitude to the hypotenuse** is $ 4\sqrt{2} $. Let's proceed step by step using geometric and algebraic reasoning.

---

### Step 1: Understand the Triangle

An **isosceles right triangle** has:
- Two equal legs of length $ a $,
- A hypotenuse of length $ a\sqrt{2} $ (by the Pythagorean Theorem),
- A right angle between the two equal legs.

The **altitude to the hypotenuse** is the perpendicular segment from the right angle to the hypotenuse. In this triangle, this altitude also acts as a **median** and **angle bisector**, and it divides the triangle into two smaller, congruent right triangles.

---

### Step 2: Use the Altitude to Find the Side Length

Let’s denote the legs of the triangle as $ a $. Then the hypotenuse is $ a\sqrt{2} $.

In any right triangle, the **altitude to the hypotenuse** is given by the formula:

$$
\text{Altitude} = \frac{\text{Product of the legs}}{\text{Hypotenuse}} = \frac{a \cdot a}{a\sqrt{2}} = \frac{a}{\sqrt{2}}
$$

We are told that this altitude is $ 4\sqrt{2} $, so:

$$
\frac{a}{\sqrt{2}} = 4\sqrt{2}
$$

Multiply both sides by $ \sqrt{2} $:

$$
a = 4\sqrt{2} \cdot \sqrt{2} = 4 \cdot 2 = 8
$$

So, the legs of the triangle are each $ 8 $ units long.

---

### Step 3: Compute the Area

The area of a triangle is given by:

$$
\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}
$$

Using the two equal legs as the base and height:

$$
\text{Area} = \frac{1}{2} \cdot a \cdot a = \frac{1}{2} \cdot 8 \cdot 8 = \frac{64}{2} = 32
$$

---

### Step 4: Verify Using the Hypotenuse and Altitude

We can also compute the area using the hypotenuse and the given altitude:

- Hypotenuse = $ a\sqrt{2} = 8\sqrt{2} $
- Altitude = $ 4\sqrt{2} $

$$
\text{Area} = \frac{1}{2} \cdot \text{hypotenuse} \cdot \text{altitude} = \frac{1}{2} \cdot 8\sqrt{2} \cdot 4\sqrt{2}
$$

$$
= \frac{1}{2} \cdot 32 \cdot 2 = \frac{64}{2} = 32
$$

This confirms the area is consistent.

---

### Final Answer

$$
\boxed{32}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of an isosceles right triangle where the altitude to the hypotenuse is given as 4√2. Hmm, let me think. First, let me recall some properties of isosceles right triangles.

An isosceles right triangle has two legs that are equal in length and a hypotenuse. The angles are 45°, 45°, and 90°. The area of a triangle is generally (base × height)/2. But here, they mention the altitude to the hypotenuse. So maybe I need to relate the altitude to the hypotenuse with the area?

Wait, let me remember. In any triangle, the area can be calculated using any side as the base, with the corresponding height. So if I take the hypotenuse as the base, then the altitude to the hypotenuse would be the height. Therefore, the area is (hypotenuse × altitude)/2. But I also know that the area can be calculated using the legs. Since it's an isosceles right triangle, the legs are equal. Let me denote the legs as 'a'. Then the hypotenuse would be a√2, right? Because in a right triangle, hypotenuse squared is equal to the sum of the squares of the legs. So, hypotenuse = √(a² + a²) = √(2a²) = a√2.

So, the area can be calculated in two ways: one using the legs and one using the hypotenuse and the altitude. Let me write that down.

First, area using legs: (a × a)/2 = a²/2.

Second, area using hypotenuse and altitude: (hypotenuse × altitude)/2 = (a√2 × 4√2)/2. Let me compute that.

Multiplying a√2 by 4√2: a * 4 * √2 * √2 = 4a * (√2 * √2) = 4a * 2 = 8a. Then divide by 2: 8a / 2 = 4a. So the area is 4a.

But we also have the area as a²/2. Therefore, setting them equal: a²/2 = 4a. Let me solve for a. Multiply both sides by 2: a² = 8a. Then subtract 8a: a² - 8a = 0. Factor: a(a - 8) = 0. So, a = 0 or a = 8. Since a is a length, it can't be zero, so a = 8.

Therefore, the legs are 8 each. Then the area is a²/2 = 8² / 2 = 64 / 2 = 32. So the area is 32? Wait, but let me check with the other formula. If hypotenuse is a√2 = 8√2, and altitude is 4√2, then area is (8√2 × 4√2)/2. Let me compute that again. 8√2 × 4√2 = 32 × (√2 × √2) = 32 × 2 = 64. Then divide by 2: 64 / 2 = 32. Yep, same result. So that seems consistent.

But let me think again if there's another way to approach this problem, just to verify. Maybe using properties of the altitude in an isosceles right triangle.

In an isosceles right triangle, the altitude to the hypotenuse also acts as a median and an angle bisector. Wait, is that true? In an isosceles triangle, the altitude from the right angle to the hypotenuse divides the triangle into two smaller, congruent triangles. Also, in this case, since it's a right triangle, the altitude to the hypotenuse can be calculated using some formula?

Alternatively, maybe I can recall that in a right triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. Let me check that.

Yes, in any right triangle, the altitude h to the hypotenuse is given by h = (leg1 × leg2)/hypotenuse.

So in this case, since it's an isosceles right triangle, leg1 = leg2 = a. Hypotenuse is a√2. Therefore, h = (a × a)/ (a√2) ) = a² / (a√2) ) = a / √2.

But the problem says that the altitude is 4√2. Therefore, a / √2 = 4√2. Let me solve for a. Multiply both sides by √2: a = 4√2 × √2 = 4 × 2 = 8. So again, a = 8. Then area is (a²)/2 = 64 / 2 = 32. Same answer.

So that seems to confirm it. Therefore, the area is 32.

But let me think again if there's another way someone might approach this problem, maybe leading to a different answer? Let me check.

Alternatively, suppose I didn't remember the formula for the altitude in a right triangle. Let me try to derive it.

Let me denote the triangle ABC, with right angle at C, and legs AC and BC equal. Let me call the hypotenuse AB. The altitude from C to AB is CD, where D is the foot of the altitude. Then, in triangle ABC, area is (AC × BC)/2 = (a²)/2. Also, area is (AB × CD)/2. Therefore, (AB × CD)/2 = (a²)/2. Therefore, AB × CD = a². Since AB = a√2, then CD = a² / AB = a² / (a√2) = a / √2. Which is the same as before. So CD = a / √2. Given CD = 4√2, so a / √2 = 4√2 => a = 4√2 × √2 = 4 × 2 = 8. So again, same result.

Therefore, area is 32.

Alternatively, maybe using coordinate geometry. Let me try that. Let me place the triangle on coordinate axes. Let me put point C at (0, 0), point A at (a, 0), and point B at (0, a). Then hypotenuse AB goes from (a, 0) to (0, a). The equation of AB is x + y = a. The altitude from C(0,0) to AB. Wait, but in this case, the altitude from C to AB is the same as the line from C to D, where D is the foot of the altitude.

But since ABC is a right triangle, the altitude from the right angle to hypotenuse is CD. Wait, but in coordinate terms, how do we find the length of CD?

Alternatively, since we know coordinates of A, B, and C, we can compute the equation of AB and then find the distance from C to AB.

Equation of AB: as above, x + y = a. The distance from point (0, 0) to the line x + y - a = 0 is |0 + 0 - a| / √(1² + 1²) ) = | -a | / √2 = a / √2. Which is the same as before. Therefore, the altitude is a / √2 = 4√2. Therefore, solving for a gives a = 8. Then area is (a²)/2 = 32.

Same answer again. So seems like no matter how I approach it, the area is 32.

But let me check once more. Maybe I made a mistake in assuming the altitude formula? Let me think.

Alternatively, suppose I didn't remember any formulas. Let me try to find the altitude.

In triangle ABC, right-angled at C, with AC = BC = a. Let me find the length of CD, where D is the foot of the altitude from C to AB.

In a right triangle, the altitude to hypotenuse divides the triangle into two smaller similar triangles. So triangle ABC ~ triangle ACD ~ triangle CBD.

Therefore, ratios of sides can be used.

In particular, AC / AB = AD / AC. Because triangle ABC ~ triangle ACD.

So AC / AB = AD / AC. Therefore, AC² = AB × AD.

Similarly, since AC = a, AB = a√2, then a² = a√2 × AD => AD = a² / (a√2) ) = a / √2.

Similarly, BD = BC² / AB = a² / (a√2) ) = a / √2. Therefore, AD = BD, which makes sense because the triangle is isosceles, so the altitude splits the hypotenuse into two equal parts. Therefore, AD = BD = a / √2. Therefore, AB = AD + DB = 2a / √2 = a√2, which matches the earlier result.

But how does this help me? Well, if I can find CD, which is the altitude. In triangle ACD, which is a right triangle, with legs AD and CD, and hypotenuse AC.

So, AC² = AD² + CD².

Therefore, a² = (a / √2)² + CD².

Calculating (a / √2)² = a² / 2. Therefore, a² = a² / 2 + CD² => CD² = a² - a² / 2 = a² / 2 => CD = a / √2. Which is the same as before.

So again, CD = a / √2 = 4√2 => a = 8. Then area is 32.

So all methods lead to the same answer. Therefore, I think the area is indeed 32.

But just to make sure, let me compute the hypotenuse length. If legs are 8, hypotenuse is 8√2. Then altitude is 4√2. Let me check if the area via hypotenuse and altitude is (8√2 × 4√2)/2.

Calculating numerator: 8√2 × 4√2 = 32 × (√2 × √2) = 32 × 2 = 64. Then divide by 2: 64 / 2 = 32. Which is the same as area via legs: (8 × 8)/2 = 64 / 2 = 32.

Therefore, all checks out.

Alternatively, maybe I can think of the altitude in terms of the triangle's properties. In an isosceles right triangle, the altitude to hypotenuse is half the length of the hypotenuse? Wait, no. Wait, if hypotenuse is a√2, then altitude is a / √2. Let me see if that's half the hypotenuse.

Half the hypotenuse would be (a√2)/2 = a / √2. Wait, that's exactly the altitude! Wait, so in this case, the altitude is equal to half the hypotenuse?

Wait, so altitude h = (a√2)/2. But we also have h = a / √2. Let me check if these are equal.

(a√2)/2 vs a / √2. Multiply numerator and denominator of a / √2 by √2: (a√2)/2. Yes, they are equal. Therefore, in an isosceles right triangle, the altitude to hypotenuse is equal to half the hypotenuse.

Therefore, h = (hypotenuse)/2. Therefore, if h = 4√2, then hypotenuse is 8√2. Then, since hypotenuse is a√2, then a√2 = 8√2 => a = 8. Then area is (a²)/2 = 64 / 2 = 32.

So that's another way to think about it. Since the altitude is half the hypotenuse, then hypotenuse is 8√2, so legs are 8 each, area is 32.

Therefore, regardless of the approach, the area is 32.

I think that's thorough enough. I can't see any mistakes in my reasoning. All methods converge to the same answer. So I think the area of the triangle is 32.

**Final Answer**
The area of the triangle is \boxed{32}.

To find the area of the isosceles right triangle, we are given that the **altitude to the hypotenuse** is $ 4\sqrt{2} $. Let's proceed step by step using geometric and algebraic reasoning.

---

### Step 1: Understand the Triangle

An **isosceles right triangle** has:
- Two equal legs of length $ a $,
- A hypotenuse of length $ a\sqrt{2} $ (by the Pythagorean Theorem),
- A right angle between the two equal legs.

The **altitude to the hypotenuse** is the perpendicular segment from the right angle to the hypotenuse. In this triangle, this altitude also acts as a **median** and **angle bisector**, and it divides the triangle into two smaller, congruent right triangles.

---

### Step 2: Use the Altitude to Find the Side Length

Let’s denote the legs of the triangle as $ a $. Then the hypotenuse is $ a\sqrt{2} $.

In any right triangle, the **altitude to the hypotenuse** is given by the formula:

$$
\text{Altitude} = \frac{\text{Product of the legs}}{\text{Hypotenuse}} = \frac{a \cdot a}{a\sqrt{2}} = \frac{a}{\sqrt{2}}
$$

We are told that this altitude is $ 4\sqrt{2} $, so:

$$
\frac{a}{\sqrt{2}} = 4\sqrt{2}
$$

Multiply both sides by $ \sqrt{2} $:

$$
a = 4\sqrt{2} \cdot \sqrt{2} = 4 \cdot 2 = 8
$$

So, the legs of the triangle are each $ 8 $ units long.

---

### Step 3: Compute the Area

The area of a triangle is given by:

$$
\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}
$$

Using the two equal legs as the base and height:

$$
\text{Area} = \frac{1}{2} \cdot a \cdot a = \frac{1}{2} \cdot 8 \cdot 8 = \frac{64}{2} = 32
$$

---

### Step 4: Verify Using the Hypotenuse and Altitude

We can also compute the area using the hypotenuse and the given altitude:

- Hypotenuse = $ a\sqrt{2} = 8\sqrt{2} $
- Altitude = $ 4\sqrt{2} $

$$
\text{Area} = \frac{1}{2} \cdot \text{hypotenuse} \cdot \text{altitude} = \frac{1}{2} \cdot 8\sqrt{2} \cdot 4\sqrt{2}
$$

$$
= \frac{1}{2} \cdot 32 \cdot 2 = \frac{64}{2} = 32
$$

This confirms the area is consistent.

---

### Final Answer

$$
\boxed{32}
$$

Solution:
In isosceles right triangle $\triangle ABC$ below, $\overline{AD}$ is the altitude to the hypotenuse.

[asy]
import olympiad;
unitsize(0.8inch);
pair A,B,C,D;
A = (0,1);
B= (1,0);
C = -B;
D = (0,0);
draw(A--B--C--A,linewidth(1));
draw(A--D,linewidth(0.8));
draw(rightanglemark(C,A,B,s=5));
draw(rightanglemark(C,D,A,s=5));
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,S);
[/asy]

Because $\triangle ABC$ is an isosceles right triangle, $\angle ABC = 45^\circ$.  Since $\angle ADB = 90^\circ$, we know that $\angle DAB = 45^\circ$, so $\triangle ABD$ is also a 45-45-90 triangle.  Similarly, $\triangle ACD$ is a 45-45-90 triangle.  Therefore, $DB=DC = DA = 4\sqrt{2}$, so $BC = BD+DC = 8\sqrt{2}$, and  \[[ABC] = \frac{(AD)(BC)}{2} = \frac{(4\sqrt{2})(8\sqrt{2})}{2} = \boxed{32}.\]


Explanation:
Okay, so I need to find the domain of the function f(x) = (2x - 7)/√(x² - 5x + 6). Hmm, let me recall what the domain of a function is. The domain is all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive because the square root of a negative number isn't a real number, and also the denominator can't be zero because division by zero is undefined.

So, first, let me focus on the denominator: √(x² - 5x + 6). For the square root to be defined (i.e., to have a real value), the expression inside the square root, which is x² - 5x + 6, must be greater than or equal to zero. However, since this is in the denominator, it can't be zero because division by zero is undefined. Therefore, the expression inside the square root must be strictly greater than zero.

So, the first step is to solve the inequality x² - 5x + 6 > 0. Let me factor the quadratic expression. Let me see, x² - 5x + 6. To factor this, I need two numbers that multiply to 6 and add up to -5. Let me think... The factors of 6 are 1 and 6, 2 and 3. Let me check: -2 and -3 multiply to 6 and add up to -5. Yes! So, x² - 5x + 6 factors into (x - 2)(x - 3).

Therefore, the inequality becomes (x - 2)(x - 3) > 0. Now, I need to solve this inequality. To do this, I can use a sign chart or test intervals. The critical points are x = 2 and x = 3. These divide the real number line into three intervals:

1. x < 2
2. 2 < x < 3
3. x > 3

I need to determine the sign of the product (x - 2)(x - 3) in each interval.

Let me pick a test point from each interval.

For the first interval, let's choose x = 1. Plugging into (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive. So, the product is positive here.

For the second interval, let's choose x = 2.5. Plugging in: (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative. So, the product is negative here.

For the third interval, let's choose x = 4. Plugging in: (4 - 2)(4 - 3) = (2)(1) = 2, which is positive. So, the product is positive here.

Therefore, the inequality (x - 2)(x - 3) > 0 is satisfied when x < 2 or x > 3.

But wait, the original expression inside the square root must be greater than zero, so the domain of the denominator is x < 2 or x > 3. However, I need to check if there are any other restrictions from the numerator. The numerator is 2x - 7, which is a linear function. It's defined for all real numbers, so there are no restrictions from the numerator.

Therefore, the domain of the function f(x) is all real numbers x such that x < 2 or x > 3.

But let me double-check. Let me make sure that there isn't any other consideration. For example, sometimes when you have functions with denominators, you have to ensure that the denominator isn't zero. But in this case, we already considered that the denominator is √(x² - 5x + 6), which is zero when x² - 5x + 6 = 0, which is exactly at x = 2 and x = 3. Therefore, those points are excluded from the domain.

So, combining all that, the domain is (-∞, 2) ∪ (3, ∞).

Wait, but let me check if there's a possibility that the numerator could be zero? For example, if the numerator is zero, does that affect the domain? Well, the domain is about where the function is defined, not about where it's zero. Even if the numerator is zero, as long as the denominator isn't zero, the function is defined. So, for example, if x = 7/2, which is 3.5, then the numerator is 2*(3.5) - 7 = 7 - 7 = 0, but the denominator is √((3.5)^2 - 5*(3.5) + 6). Let me compute that: 3.5 squared is 12.25, 5*3.5 is 17.5, so 12.25 - 17.5 + 6 = 12.25 - 17.5 is -5.25 + 6 is 0.75. So, √0.75 is defined and non-zero. Therefore, x = 3.5 is in the domain. So, even if the numerator is zero, as long as the denominator is not zero, it's okay. Therefore, the domain is indeed x < 2 or x > 3.

But let me check another point in the interval between 2 and 3. Let's say x = 2.5. Then, the denominator is √( (2.5)^2 - 5*(2.5) + 6 ). Let me compute that: 6.25 - 12.5 + 6 = (6.25 + 6) - 12.5 = 12.25 - 12.5 = -0.25. So, square root of a negative number, which is not real. Therefore, x = 2.5 is not in the domain, which matches our previous conclusion.

Therefore, the domain is all real numbers except those between 2 and 3, including 2 and 3? Wait, no. Wait, the inequality was x² - 5x + 6 > 0, so x < 2 or x > 3. Therefore, 2 and 3 are excluded. So, the domain is (-∞, 2) ∪ (3, ∞).

Therefore, the answer should be all real numbers x such that x is less than 2 or greater than 3.

But let me check once more. Let me think if there's any other possible mistake. For example, sometimes when you have square roots, you have to consider if the expression inside is non-negative, but since it's in the denominator, it must be strictly positive. So, yes, the expression inside must be greater than zero, not just greater than or equal. Therefore, the solution is correct.

Alternatively, if the function was f(x) = (2x - 7)/√(x² - 5x + 6), then the denominator is √(quadratic). So, the quadratic must be positive. Therefore, the domain is x in (-∞, 2) ∪ (3, ∞).

Therefore, I think that's the correct answer.

**Final Answer**
The domain of the function is \boxed{(-\infty, 2) \cup (3, \infty)}.

To determine the **domain** of the function
$$
f(x) = \frac{2x - 7}{\sqrt{x^2 - 5x + 6}},
$$
we need to identify all real values of $ x $ for which the function is defined.

---

### Step 1: Understand the Structure of the Function

The function has a **rational expression** with a **square root in the denominator**. This means:

- The **numerator** $ 2x - 7 $ is a linear expression and is defined for all real numbers.
- The **denominator** $ \sqrt{x^2 - 5x + 6} $ introduces two key restrictions:
  1. The expression inside the square root must be **non-negative** (i.e., $ x^2 - 5x + 6 \geq 0 $).
  2. The square root must be **non-zero** (i.e., $ x^2 - 5x + 6 > 0 $), because division by zero is undefined.

So, the **denominator must be strictly positive**.

---

### Step 2: Solve the Inequality

We solve the inequality:
$$
x^2 - 5x + 6 > 0
$$

#### Factor the quadratic:
$$
x^2 - 5x + 6 = (x - 2)(x - 3)
$$

So the inequality becomes:
$$
(x - 2)(x - 3) > 0
$$

#### Determine the sign of the product:

- The critical points are $ x = 2 $ and $ x = 3 $.
- These divide the number line into three intervals:
  - $ x < 2 $
  - $ 2 < x < 3 $
  - $ x > 3 $

We test the sign of the product in each interval:

- For $ x < 2 $, say $ x = 1 $: $ (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0 $
- For $ 2 < x < 3 $, say $ x = 2.5 $: $ (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 $
- For $ x > 3 $, say $ x = 4 $: $ (4 - 2)(4 - 3) = (2)(1) = 2 > 0 $

So the inequality $ (x - 2)(x - 3) > 0 $ is satisfied when:
$$
x < 2 \quad \text{or} \quad x > 3
$$

---

### Step 3: Final Domain

Since the denominator must be **strictly positive**, the domain of the function is:
$$
x \in (-\infty, 2) \cup (3, \infty)
$$

There are no additional restrictions from the numerator, as it is defined for all real numbers.

---

### Final Answer

$$
\boxed{(-\infty, 2) \cup (3, \infty)}
$$

[Assistant]
Explanation:
Okay, so I need to find the domain of the function f(x) = (2x - 7)/√(x² - 5x + 6). Hmm, let me recall what the domain of a function is. The domain is all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive because the square root of a negative number isn't a real number, and also the denominator can't be zero because division by zero is undefined.

So, first, let me focus on the denominator: √(x² - 5x + 6). For the square root to be defined (i.e., to have a real value), the expression inside the square root, which is x² - 5x + 6, must be greater than or equal to zero. However, since this is in the denominator, it can't be zero because division by zero is undefined. Therefore, the expression inside the square root must be strictly greater than zero.

So, the first step is to solve the inequality x² - 5x + 6 > 0. Let me factor the quadratic expression. Let me see, x² - 5x + 6. To factor this, I need two numbers that multiply to 6 and add up to -5. Let me think... The factors of 6 are 1 and 6, 2 and 3. Let me check: -2 and -3 multiply to 6 and add up to -5. Yes! So, x² - 5x + 6 factors into (x - 2)(x - 3).

Therefore, the inequality becomes (x - 2)(x - 3) > 0. Now, I need to solve this inequality. To do this, I can use a sign chart or test intervals. The critical points are x = 2 and x = 3. These divide the real number line into three intervals:

1. x < 2
2. 2 < x < 3
3. x > 3

I need to determine the sign of the product (x - 2)(x - 3) in each interval.

Let me pick a test point from each interval.

For the first interval, let's choose x = 1. Plugging into (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive. So, the product is positive here.

For the second interval, let's choose x = 2.5. Plugging in: (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative. So, the product is negative here.

For the third interval, let's choose x = 4. Plugging in: (4 - 2)(4 - 3) = (2)(1) = 2, which is positive. So, the product is positive here.

Therefore, the inequality (x - 2)(x - 3) > 0 is satisfied when x < 2 or x > 3.

But wait, the original expression inside the square root must be greater than zero, so the domain of the denominator is x < 2 or x > 3. However, I need to check if there are any other restrictions from the numerator. The numerator is 2x - 7, which is a linear function. It's defined for all real numbers, so there are no restrictions from the numerator.

Therefore, the domain of the function f(x) is all real numbers x such that x < 2 or x > 3.

But let me double-check. Let me make sure that there isn't any other consideration. For example, sometimes when you have functions with denominators, you have to ensure that the denominator isn't zero. But in this case, we already considered that the denominator is √(x² - 5x + 6), which is zero when x² - 5x + 6 = 0, which is exactly at x = 2 and x = 3. Therefore, those points are excluded from the domain.

So, combining all that, the domain is (-∞, 2) ∪ (3, ∞).

Wait, but let me check if there's a possibility that the numerator could be zero? For example, if the numerator is zero, does that affect the domain? Well, the domain is about where the function is defined, not about where it's zero. Even if the numerator is zero, as long as the denominator isn't zero, the function is defined. So, for example, if x = 7/2, which is 3.5, then the numerator is 2*(3.5) - 7 = 7 - 7 = 0, but the denominator is √((3.5)^2 - 5*(3.5) + 6). Let me compute that: 3.5 squared is 12.25, 5*3.5 is 17.5, so 12.25 - 17.5 + 6 = 12.25 - 17.5 is -5.25 + 6 is 0.75. So, √0.75 is defined and non-zero. Therefore, x = 3.5 is in the domain. So, even if the numerator is zero, as long as the denominator is not zero, it's okay. Therefore, the domain is indeed x < 2 or x > 3.

But let me check another point in the interval between 2 and 3. Let's say x = 2.5. Then, the denominator is √( (2.5)^2 - 5*(2.5) + 6 ). Let me compute that: 6.25 - 12.5 + 6 = (6.25 + 6) - 12.5 = 12.25 - 12.5 = -0.25. So, square root of a negative number, which is not real. Therefore, x = 2.5 is not in the domain, which matches our previous conclusion.

Therefore, the domain is all real numbers except those between 2 and 3, including 2 and 3? Wait, no. Wait, the inequality was x² - 5x + 6 > 0, so x < 2 or x > 3. Therefore, 2 and 3 are excluded. So, the domain is (-∞, 2) ∪ (3, ∞).

Therefore, the answer should be all real numbers x such that x is less than 2 or greater than 3.

But let me check once more. Let me think if there's any other possible mistake. For example, sometimes when you have square roots, you have to consider if the expression inside is non-negative, but since it's in the denominator, it must be strictly positive. So, yes, the expression inside must be greater than zero, not just greater than or equal. Therefore, the solution is correct.

Alternatively, if the function was f(x) = (2x - 7)/√(x² - 5x + 6), then the denominator is √(quadratic). So, the quadratic must be positive. Therefore, the domain is x in (-∞, 2) ∪ (3, ∞).

Therefore, I think that's the correct answer.

**Final Answer**
The domain of the function is \boxed{(-\infty, 2) \cup (3, \infty)}.

To determine the **domain** of the function
$$
f(x) = \frac{2x - 7}{\sqrt{x^2 - 5x + 6}},
$$
we need to identify all real values of $ x $ for which the function is defined.

---

### Step 1: Understand the Structure of the Function

The function has a **rational expression** with a **square root in the denominator**. This means:

- The **numerator** $ 2x - 7 $ is a linear expression and is defined for all real numbers.
- The **denominator** $ \sqrt{x^2 - 5x + 6} $ introduces two key restrictions:
  1. The expression inside the square root must be **non-negative** (i.e., $ x^2 - 5x + 6 \geq 0 $).
  2. The square root must be **non-zero** (i.e., $ x^2 - 5x + 6 > 0 $), because division by zero is undefined.

So, the **denominator must be strictly positive**.

---

### Step 2: Solve the Inequality

We solve the inequality:
$$
x^2 - 5x + 6 > 0
$$

#### Factor the quadratic:
$$
x^2 - 5x + 6 = (x - 2)(x - 3)
$$

So the inequality becomes:
$$
(x - 2)(x - 3) > 0
$$

#### Determine the sign of the product:

- The critical points are $ x = 2 $ and $ x = 3 $.
- These divide the number line into three intervals:
  - $ x < 2 $
  - $ 2 < x < 3 $
  - $ x > 3 $

We test the sign of the product in each interval:

- For $ x < 2 $, say $ x = 1 $: $ (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0 $
- For $ 2 < x < 3 $, say $ x = 2.5 $: $ (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 $
- For $ x > 3 $, say $ x = 4 $: $ (4 - 2)(4 - 3) = (2)(1) = 2 > 0 $

So the inequality $ (x - 2)(x - 3) > 0 $ is satisfied when:
$$
x < 2 \quad \text{or} \quad x > 3
$$

---

### Step 3: Final Domain

Since the denominator must be **strictly positive**, the domain of the function is:
$$
x \in (-\infty, 2) \cup (3, \infty)
$$

There are no additional restrictions from the numerator, as it is defined for all real numbers.

---

### Final Answer

$$
\boxed{(-\infty, 2) \cup (3, \infty)}
$$

Solution: The function is defined when the value inside the square root is positive, i.e. we must have $x^2-5x+6>0$.  Factoring, we get $(x-3)(x-2)>0$. So either both factors in the left hand side are negative or they are both positive. They are both negative when $x<2$. They are both positive when $x>3$. So the domain of $f(x)$ is $x<2 \text{ or } x>3$, or $x \in \boxed{(-\infty, 2) \cup (3, \infty)}$ in interval notation.