Explanation:
Okay, let's try to figure out this problem. So, there's a group of N students, and N is less than 50. The teacher tried grouping them in groups of 8, and the last group had 5 students. Then, if they tried grouping them in groups of 6, the last group had 3 students. We need to find all possible values of N and then add them up. Alright, let me break this down.

First, the problem mentions two different grouping scenarios. Let me translate these into mathematical equations. When the teacher divides them into groups of 8, there's a remainder of 5. That means N divided by 8 leaves a remainder of 5. Similarly, when divided by 6, the remainder is 3. So, these are congruence relations.

So, mathematically, we can write:

1. N ≡ 5 mod 8
2. N ≡ 3 mod 6

And we also know that N < 50. The question is asking for the sum of all possible N that satisfy these conditions. Alright, let me recall how to solve such problems. This seems like a system of congruences, so maybe the Chinese Remainder Theorem applies here? Although, since 8 and 6 are not coprime (their GCD is 2), the Chinese Remainder Theorem in its basic form might not apply directly. Hmm. Let me check.

But maybe I can solve the system step by step. Let me first consider the first congruence: N ≡ 5 mod 8. That means N can be written as 8k + 5, where k is some integer. Since N < 50, let me find the possible values of k.

Let me compute the maximum value of k. 8k + 5 < 50 => 8k < 45 => k < 45/8 => k < 5.625. Since k is an integer, k can be from 0 up to 5. Let me list these possible N:

For k=0: 8*0 +5 = 5
k=1: 13
k=2: 21
k=3: 29
k=4: 37
k=5: 45

So, possible N values from the first condition are 5,13,21,29,37,45.

Now, we need to check which of these also satisfy the second condition: N ≡ 3 mod 6. Let me compute each of these numbers modulo 6.

Starting with N=5: 5 divided by 6 is 0 with remainder 5. So 5 mod 6 is 5. But we need 3. Not good.

Next N=13: 13 divided by 6 is 2 with remainder 1. 13 mod 6 is 1. Not 3. Nope.

N=21: 21 divided by 6 is 3 with remainder 3. So 21 mod 6 is 3. Hey, this works!

N=29: 29 divided by 6: 6*4=24, 29-24=5. So 29 mod 6 is 5. Not 3.

N=37: 37 divided by 6: 6*6=36, remainder 1. So 37 mod 6 is 1. Not 3.

N=45: 45 divided by 6 is 7*6=42, remainder 3. So 45 mod 6 is 3. That works too!

So, from the list, N=21 and N=45 satisfy both conditions. Wait, let me check again. Wait, N=21: 21/8 is 2 groups of 8 with remainder 5? Let me confirm. 8*2=16, 21-16=5. Yes. And 21 divided by 6 is 3 groups with remainder 3. Correct. Similarly, 45 divided by 8: 8*5=40, 45-40=5. Yes, remainder 5. Divided by 6: 6*7=42, 45-42=3. Correct. So those two numbers are valid.

But wait, are there any other numbers? The possible N's from the first condition were 5,13,21,29,37,45. Only 21 and 45 satisfied the second condition. Therefore, the possible Ns are 21 and 45. The problem says N < 50, so 45 is okay because it's less than 50. Wait, but 45 is less than 50? Yes, 45 is 45, which is less than 50. So both are valid. So the answer is 21 + 45 = 66?

Wait, but let me check if there are any other numbers. Maybe I missed some? Let me think again. The first condition gives N=5,13,21,29,37,45. Let me check again each of them:

N=5: 5 mod 6=5≠3

N=13: 13 mod 6=1≠3

N=21: 21 mod 6=3 ✔

N=29: 29 mod 6=5≠3

N=37: 37 mod 6=1≠3

N=45: 45 mod 6=3 ✔

So yes, only 21 and 45. Then their sum is 21 + 45 = 66. Therefore, the answer should be 66. But wait, let me check again if there's another approach where maybe more numbers satisfy. Maybe I made an error in generating the possible Ns?

Let me try solving the system of congruences algebraically. Let me write:

From the first equation, N ≡ 5 mod 8 => N = 8k +5.

Substitute into the second equation: 8k +5 ≡ 3 mod 6 => 8k ≡ -2 mod 6 => 8k ≡ 4 mod 6.

But since 8 ≡ 2 mod 6, this becomes 2k ≡ 4 mod 6. Let me solve 2k ≡ 4 mod 6.

Divide both sides by 2: k ≡ 2 mod 3. So k = 3m + 2, where m is an integer. Therefore, substituting back into N:

N = 8k +5 = 8*(3m +2) +5 = 24m +16 +5 = 24m +21.

Therefore, all solutions are N ≡21 mod 24. So N=21, 21+24=45, 45+24=69, etc. Since N <50, possible values are 21 and 45. Hence, sum is 21+45=66. So that confirms my earlier result. So the answer is 66. Therefore, the sum of all possible values of N is 66. Wait, but let me check once again.

Alternative approach: Let me think of numbers less than 50 that satisfy N ≡5 mod 8 and N≡3 mod6. Let me check numbers congruent to 5 mod 8 first.

Numbers congruent to 5 mod 8 under 50 are as I listed before: 5,13,21,29,37,45. Then check which of these are congruent to 3 mod6. As above, 21 and 45. So same result. So, answer is 21+45=66. Therefore, I think that is correct.

Wait, but just to make sure, let me check if there's another way someone might think. For example, maybe by finding numbers that are 3 mod6, and then check if they are 5 mod8. Let me try that approach.

Numbers less than 50 that are 3 mod6: Starting from 3: 3, 9, 15, 21, 27, 33, 39, 45, 51... but 51 is over 50. So numbers are 3,9,15,21,27,33,39,45. Now check which of these are 5 mod8.

Check each:

3: 3 mod8=3≠5. No.

9: 9 mod8=1≠5. No.

15: 15 mod8=7≠5. No.

21: 21 mod8=5. Yes! 21 is good.

27: 27 mod8=3 (since 8*3=24, 27-24=3). No.

33: 33 mod8: 8*4=32, 33-32=1. No.

39: 39 mod8: 8*4=32, 39-32=7. No.

45: 45 mod8: 8*5=40, 45-40=5. Yes! So again, 21 and 45. Same numbers. Therefore, same answer. So sum is 21+45=66. Therefore, I think this is correct.

Wait, but just to check again, maybe there's something wrong here? Let me check the initial equations again. If N ≡5 mod8, so when divided by 8, remainder 5. So N=8k+5. Then if we divide that by 6, remainder 3. So 8k +5 ≡3 mod6. Which is 8k ≡ -2 mod6. 8k ≡4 mod6. Then since 8≡2 mod6, so 2k≡4 mod6. Dividing both sides by 2 gives k≡2 mod3. Therefore k=3m +2. Then N=8*(3m +2)+5=24m +21. So N=21, 45, 69,... Since N <50, only 21 and 45. So yes. So that's the same result.

Therefore, I think the answer is 66. So the sum of all possible values of N is 21 + 45 = 66.

But let me check once again if there's a possible mistake in the modulus operations. For example, when solving 2k ≡4 mod6, is k ≡2 mod3? Let me verify.

The equation 2k ≡4 mod6 can be simplified by dividing by 2, but we have to be careful because 2 and 6 are not coprime. The equation 2k ≡4 mod6 is equivalent to 2k -4 is divisible by 6, i.e., 2(k - 2) is divisible by 6, which implies that k - 2 must be divisible by 3. Therefore, k ≡2 mod3. Therefore, k=3m +2. Hence, yes, correct. Therefore, N=24m +21. Hence, only 21 and 45 under 50. So answer is 66.

Another way to think: Let me think of numbers that satisfy both conditions. For example, numbers that are 5 more than a multiple of 8, and 3 more than a multiple of 6. So, the difference between N and 5 is a multiple of 8, and the difference between N and 3 is a multiple of 6. Therefore, N -5 is divisible by 8, N -3 is divisible by 6. Therefore, N -5 =8k, N -3=6m. Therefore, 8k = N -5, 6m = N -3. Therefore, subtract these equations: 6m -8k = (N -3) - (N -5) = 2. Therefore, 6m -8k=2. Simplify: divide both sides by 2: 3m -4k=1. So, solving for integers m and k: 3m -4k=1. Let me solve this Diophantine equation.

We can rearrange it as 3m = 4k +1. So 3m ≡1 mod4. Therefore, m ≡(1/3) mod4. Since 3*3=9≡1 mod4, so inverse of 3 mod4 is 3. Therefore, m ≡3*1≡3 mod4. Therefore, m=4t +3 for some integer t. Therefore, substituting back into 3m=4k +1: 3*(4t +3)=4k +1 => 12t +9=4k +1 => 4k=12t +8 => k=3t +2. Therefore, k=3t +2. Therefore, N -5=8k=8*(3t +2)=24t +16. Therefore, N=24t +16 +5=24t +21. Therefore, N=24t +21. Hence, general solution. Therefore, N=21, 45, 69,... Since N <50, t=0: 21, t=1: 45, t=2: 69 which is over 50. Therefore, only 21 and 45. Same result. So sum is 66.

Therefore, regardless of the method used, the answer seems to be 66. Therefore, I think that's the correct answer.

But just to make sure there are no other numbers, perhaps I can list all numbers less than 50 that satisfy both conditions. Let me try to list them.

First, numbers that are 5 mod8: 5,13,21,29,37,45. Check which of those are 3 mod6. As before:

5 mod6=5≠3; 13 mod6=1≠3; 21 mod6=3 ✔; 29 mod6=5≠3; 37 mod6=1≠3; 45 mod6=3 ✔. So only 21 and 45. Therefore, sum is 21 + 45=66. Yep, same.

So, I think I can be confident that the answer is 66.

**Final Answer**
\boxed{66}

To solve the problem, we are given two conditions about a group of $ N < 50 $ students:

1. If grouped into groups of 8, the last group has 5 students.
2. If grouped into groups of 6, the last group has 3 students.

---

### Step 1: Interpreting as Congruences

These statements can be translated into modular arithmetic:
- $ N \equiv 5 \pmod{8} $
- $ N \equiv 3 \pmod{6} $

We are to find all $ N < 50 $ that satisfy both of these congruences and then sum the valid values of $ N $.

---

### Step 2: Solve the System of Congruences

We can solve the system:
$$
\begin{cases}
N \equiv 5 \pmod{8} \\
N \equiv 3 \pmod{6}
\end{cases}
$$

Let’s express $ N $ in terms of the first congruence:
$$
N = 8k + 5
$$
Substitute into the second congruence:
$$
8k + 5 \equiv 3 \pmod{6}
\Rightarrow 8k \equiv -2 \pmod{6}
\Rightarrow 8k \equiv 4 \pmod{6}
$$

Now reduce modulo 6:
$$
8 \equiv 2 \pmod{6} \Rightarrow 2k \equiv 4 \pmod{6}
$$

Divide both sides by 2:
$$
k \equiv 2 \pmod{3}
\Rightarrow k = 3m + 2
$$

Substitute back to find $ N $:
$$
N = 8k + 5 = 8(3m + 2) + 5 = 24m + 21
$$

So the general solution is:
$$
N \equiv 21 \pmod{24}
\Rightarrow N = 21, 45, 69, \dots
$$

Given $ N < 50 $, the only valid values of $ N $ are:
$$
N = 21 \text{ and } N = 45
$$

---

### Step 3: Compute the Sum

Now, summing the valid values:
$$
21 + 45 = \boxed{66}
$$

---

### ✅ Final Answer

$$
\boxed{66}
$$

[Assistant]
Explanation:
Okay, let's try to figure out this problem. So, there's a group of N students, and N is less than 50. The teacher tried grouping them in groups of 8, and the last group had 5 students. Then, if they tried grouping them in groups of 6, the last group had 3 students. We need to find all possible values of N and then add them up. Alright, let me break this down.

First, the problem mentions two different grouping scenarios. Let me translate these into mathematical equations. When the teacher divides them into groups of 8, there's a remainder of 5. That means N divided by 8 leaves a remainder of 5. Similarly, when divided by 6, the remainder is 3. So, these are congruence relations.

So, mathematically, we can write:

1. N ≡ 5 mod 8
2. N ≡ 3 mod 6

And we also know that N < 50. The question is asking for the sum of all possible N that satisfy these conditions. Alright, let me recall how to solve such problems. This seems like a system of congruences, so maybe the Chinese Remainder Theorem applies here? Although, since 8 and 6 are not coprime (their GCD is 2), the Chinese Remainder Theorem in its basic form might not apply directly. Hmm. Let me check.

But maybe I can solve the system step by step. Let me first consider the first congruence: N ≡ 5 mod 8. That means N can be written as 8k + 5, where k is some integer. Since N < 50, let me find the possible values of k.

Let me compute the maximum value of k. 8k + 5 < 50 => 8k < 45 => k < 45/8 => k < 5.625. Since k is an integer, k can be from 0 up to 5. Let me list these possible N:

For k=0: 8*0 +5 = 5
k=1: 13
k=2: 21
k=3: 29
k=4: 37
k=5: 45

So, possible N values from the first condition are 5,13,21,29,37,45.

Now, we need to check which of these also satisfy the second condition: N ≡ 3 mod 6. Let me compute each of these numbers modulo 6.

Starting with N=5: 5 divided by 6 is 0 with remainder 5. So 5 mod 6 is 5. But we need 3. Not good.

Next N=13: 13 divided by 6 is 2 with remainder 1. 13 mod 6 is 1. Not 3. Nope.

N=21: 21 divided by 6 is 3 with remainder 3. So 21 mod 6 is 3. Hey, this works!

N=29: 29 divided by 6: 6*4=24, 29-24=5. So 29 mod 6 is 5. Not 3.

N=37: 37 divided by 6: 6*6=36, remainder 1. So 37 mod 6 is 1. Not 3.

N=45: 45 divided by 6 is 7*6=42, remainder 3. So 45 mod 6 is 3. That works too!

So, from the list, N=21 and N=45 satisfy both conditions. Wait, let me check again. Wait, N=21: 21/8 is 2 groups of 8 with remainder 5? Let me confirm. 8*2=16, 21-16=5. Yes. And 21 divided by 6 is 3 groups with remainder 3. Correct. Similarly, 45 divided by 8: 8*5=40, 45-40=5. Yes, remainder 5. Divided by 6: 6*7=42, 45-42=3. Correct. So those two numbers are valid.

But wait, are there any other numbers? The possible N's from the first condition were 5,13,21,29,37,45. Only 21 and 45 satisfied the second condition. Therefore, the possible Ns are 21 and 45. The problem says N < 50, so 45 is okay because it's less than 50. Wait, but 45 is less than 50? Yes, 45 is 45, which is less than 50. So both are valid. So the answer is 21 + 45 = 66?

Wait, but let me check if there are any other numbers. Maybe I missed some? Let me think again. The first condition gives N=5,13,21,29,37,45. Let me check again each of them:

N=5: 5 mod 6=5≠3

N=13: 13 mod 6=1≠3

N=21: 21 mod 6=3 ✔

N=29: 29 mod 6=5≠3

N=37: 37 mod 6=1≠3

N=45: 45 mod 6=3 ✔

So yes, only 21 and 45. Then their sum is 21 + 45 = 66. Therefore, the answer should be 66. But wait, let me check again if there's another approach where maybe more numbers satisfy. Maybe I made an error in generating the possible Ns?

Let me try solving the system of congruences algebraically. Let me write:

From the first equation, N ≡ 5 mod 8 => N = 8k +5.

Substitute into the second equation: 8k +5 ≡ 3 mod 6 => 8k ≡ -2 mod 6 => 8k ≡ 4 mod 6.

But since 8 ≡ 2 mod 6, this becomes 2k ≡ 4 mod 6. Let me solve 2k ≡ 4 mod 6.

Divide both sides by 2: k ≡ 2 mod 3. So k = 3m + 2, where m is an integer. Therefore, substituting back into N:

N = 8k +5 = 8*(3m +2) +5 = 24m +16 +5 = 24m +21.

Therefore, all solutions are N ≡21 mod 24. So N=21, 21+24=45, 45+24=69, etc. Since N <50, possible values are 21 and 45. Hence, sum is 21+45=66. So that confirms my earlier result. So the answer is 66. Therefore, the sum of all possible values of N is 66. Wait, but let me check once again.

Alternative approach: Let me think of numbers less than 50 that satisfy N ≡5 mod 8 and N≡3 mod6. Let me check numbers congruent to 5 mod 8 first.

Numbers congruent to 5 mod 8 under 50 are as I listed before: 5,13,21,29,37,45. Then check which of these are congruent to 3 mod6. As above, 21 and 45. So same result. So, answer is 21+45=66. Therefore, I think that is correct.

Wait, but just to make sure, let me check if there's another way someone might think. For example, maybe by finding numbers that are 3 mod6, and then check if they are 5 mod8. Let me try that approach.

Numbers less than 50 that are 3 mod6: Starting from 3: 3, 9, 15, 21, 27, 33, 39, 45, 51... but 51 is over 50. So numbers are 3,9,15,21,27,33,39,45. Now check which of these are 5 mod8.

Check each:

3: 3 mod8=3≠5. No.

9: 9 mod8=1≠5. No.

15: 15 mod8=7≠5. No.

21: 21 mod8=5. Yes! 21 is good.

27: 27 mod8=3 (since 8*3=24, 27-24=3). No.

33: 33 mod8: 8*4=32, 33-32=1. No.

39: 39 mod8: 8*4=32, 39-32=7. No.

45: 45 mod8: 8*5=40, 45-40=5. Yes! So again, 21 and 45. Same numbers. Therefore, same answer. So sum is 21+45=66. Therefore, I think this is correct.

Wait, but just to check again, maybe there's something wrong here? Let me check the initial equations again. If N ≡5 mod8, so when divided by 8, remainder 5. So N=8k+5. Then if we divide that by 6, remainder 3. So 8k +5 ≡3 mod6. Which is 8k ≡ -2 mod6. 8k ≡4 mod6. Then since 8≡2 mod6, so 2k≡4 mod6. Dividing both sides by 2 gives k≡2 mod3. Therefore k=3m +2. Then N=8*(3m +2)+5=24m +21. So N=21, 45, 69,... Since N <50, only 21 and 45. So yes. So that's the same result.

Therefore, I think the answer is 66. So the sum of all possible values of N is 21 + 45 = 66.

But let me check once again if there's a possible mistake in the modulus operations. For example, when solving 2k ≡4 mod6, is k ≡2 mod3? Let me verify.

The equation 2k ≡4 mod6 can be simplified by dividing by 2, but we have to be careful because 2 and 6 are not coprime. The equation 2k ≡4 mod6 is equivalent to 2k -4 is divisible by 6, i.e., 2(k - 2) is divisible by 6, which implies that k - 2 must be divisible by 3. Therefore, k ≡2 mod3. Therefore, k=3m +2. Hence, yes, correct. Therefore, N=24m +21. Hence, only 21 and 45 under 50. So answer is 66.

Another way to think: Let me think of numbers that satisfy both conditions. For example, numbers that are 5 more than a multiple of 8, and 3 more than a multiple of 6. So, the difference between N and 5 is a multiple of 8, and the difference between N and 3 is a multiple of 6. Therefore, N -5 is divisible by 8, N -3 is divisible by 6. Therefore, N -5 =8k, N -3=6m. Therefore, 8k = N -5, 6m = N -3. Therefore, subtract these equations: 6m -8k = (N -3) - (N -5) = 2. Therefore, 6m -8k=2. Simplify: divide both sides by 2: 3m -4k=1. So, solving for integers m and k: 3m -4k=1. Let me solve this Diophantine equation.

We can rearrange it as 3m = 4k +1. So 3m ≡1 mod4. Therefore, m ≡(1/3) mod4. Since 3*3=9≡1 mod4, so inverse of 3 mod4 is 3. Therefore, m ≡3*1≡3 mod4. Therefore, m=4t +3 for some integer t. Therefore, substituting back into 3m=4k +1: 3*(4t +3)=4k +1 => 12t +9=4k +1 => 4k=12t +8 => k=3t +2. Therefore, k=3t +2. Therefore, N -5=8k=8*(3t +2)=24t +16. Therefore, N=24t +16 +5=24t +21. Therefore, N=24t +21. Hence, general solution. Therefore, N=21, 45, 69,... Since N <50, t=0: 21, t=1: 45, t=2: 69 which is over 50. Therefore, only 21 and 45. Same result. So sum is 66.

Therefore, regardless of the method used, the answer seems to be 66. Therefore, I think that's the correct answer.

But just to make sure there are no other numbers, perhaps I can list all numbers less than 50 that satisfy both conditions. Let me try to list them.

First, numbers that are 5 mod8: 5,13,21,29,37,45. Check which of those are 3 mod6. As before:

5 mod6=5≠3; 13 mod6=1≠3; 21 mod6=3 ✔; 29 mod6=5≠3; 37 mod6=1≠3; 45 mod6=3 ✔. So only 21 and 45. Therefore, sum is 21 + 45=66. Yep, same.

So, I think I can be confident that the answer is 66.

**Final Answer**
\boxed{66}

To solve the problem, we are given two conditions about a group of $ N < 50 $ students:

1. If grouped into groups of 8, the last group has 5 students.
2. If grouped into groups of 6, the last group has 3 students.

---

### Step 1: Interpreting as Congruences

These statements can be translated into modular arithmetic:
- $ N \equiv 5 \pmod{8} $
- $ N \equiv 3 \pmod{6} $

We are to find all $ N < 50 $ that satisfy both of these congruences and then sum the valid values of $ N $.

---

### Step 2: Solve the System of Congruences

We can solve the system:
$$
\begin{cases}
N \equiv 5 \pmod{8} \\
N \equiv 3 \pmod{6}
\end{cases}
$$

Let’s express $ N $ in terms of the first congruence:
$$
N = 8k + 5
$$
Substitute into the second congruence:
$$
8k + 5 \equiv 3 \pmod{6}
\Rightarrow 8k \equiv -2 \pmod{6}
\Rightarrow 8k \equiv 4 \pmod{6}
$$

Now reduce modulo 6:
$$
8 \equiv 2 \pmod{6} \Rightarrow 2k \equiv 4 \pmod{6}
$$

Divide both sides by 2:
$$
k \equiv 2 \pmod{3}
\Rightarrow k = 3m + 2
$$

Substitute back to find $ N $:
$$
N = 8k + 5 = 8(3m + 2) + 5 = 24m + 21
$$

So the general solution is:
$$
N \equiv 21 \pmod{24}
\Rightarrow N = 21, 45, 69, \dots
$$

Given $ N < 50 $, the only valid values of $ N $ are:
$$
N = 21 \text{ and } N = 45
$$

---

### Step 3: Compute the Sum

Now, summing the valid values:
$$
21 + 45 = \boxed{66}
$$

---

### ✅ Final Answer

$$
\boxed{66}
$$

Solution: We are given that $N\equiv 5\pmod{8}$ and $N\equiv 3\pmod{6}$.  We begin checking numbers which are 5 more than a multiple of 8, and we find that 5 and 13 are not 3 more than a multiple of 6, but 21 is 3 more than a multiple of 6. Thus 21 is one possible value of $N$. By the Chinese Remainder Theorem, the integers $x$ satisfying $x\equiv 5\pmod{8}$ and $x\equiv 3\pmod{6}$ are those of the form $x=21+\text{lcm}(6,8)k = 21 + 24 k$, where $k$ is an integer. Thus the 2 solutions less than $50$ are 21 and $21+24(1) = 45$, and their sum is $21+45=\boxed{66}$.


Explanation:
Okay, let's see. There's this Asymptote figure showing a rectangle made up of twelve 1 by 1 squares. The question is asking for the total area of the shaded region. Hmm. Let me try to visualize it based on the Asymptote code provided.

First, the Asymptote code draws a grid. It has vertical lines from j = 0 to 3 (so four vertical lines, which would make 3 units in width) and horizontal lines from j = 0 to 4 (five horizontal lines, so four units in height). Wait, but twelve 1x1 squares... So, if it's a rectangle, the number of squares would be width times height. Let me check. If the grid is 3 units wide (since there are 4 vertical lines, from x=0 to x=3) and 4 units tall (from y=0 to y=4), then the total area of the rectangle would be 3*4=12 squares. Yep, that matches the problem statement. So the figure is a 3x4 rectangle made up of twelve 1x1 squares.

Now, the shaded regions are filled with gray. The Asymptote code has two fill commands. Let me parse them:

First fill: (0,4)--(0,0)--(2,0)--cycle. So that's a triangle with vertices at (0,4), (0,0), and (2,0). So this is a right triangle with base along the y-axis from (0,4) to (0,0), which is 4 units, and then a base along the x-axis from (0,0) to (2,0), which is 2 units. Wait, but connecting those points would form a triangle? Wait, connecting (0,4) to (0,0) to (2,0) and back to (0,4)? So that's a right triangle with legs of length 4 and 2? Wait, no. Wait, if you connect (0,4) to (0,0) to (2,0), then back to (0,4), the triangle would have a base of 2 units (from (0,0) to (2,0)) and a height of 4 units? Wait, but in that case, the area would be (base * height)/2. But is that accurate?

Alternatively, maybe I should think in terms of coordinates. The triangle with vertices at (0,4), (0,0), and (2,0). Let me sketch this mentally. The vertical line from (0,0) to (0,4), then a horizontal line to (2,0), then back to (0,4). So the triangle is a right triangle with base along the x-axis from (0,0) to (2,0) and height along the y-axis from (0,0) to (0,4). But since it's a triangle, not a rectangle, the area would be ½ * base * height. So base is 2 units, height is 4 units? Wait, but in reality, the triangle is not a right triangle with legs 2 and 4. Wait, actually, if you have points (0,4), (0,0), (2,0), then the triangle has a vertical side of length 4, a horizontal side of length 2, and then the hypotenuse connecting (0,4) to (2,0). So yes, this is a right triangle with legs 4 and 2? Wait, no. Wait, actually, the legs would be the vertical side from (0,0) to (0,4) which is length 4, and the horizontal side from (0,0) to (2,0) which is length 2, so the area is ½ * 4 * 2 = 4. So that triangle has area 4?

But let me confirm with coordinates. Using the formula for area of a triangle with coordinates. The coordinates are (0,4), (0,0), (2,0). Using the shoelace formula:

List the coordinates:

(0,4), (0,0), (2,0), (0,4)

Compute sum of products going one way:

(0*0) + (0*0) + (2*4) = 0 + 0 + 8 = 8

Minus sum of products the other way:

(4*0) + (0*2) + (0*0) = 0 + 0 + 0 = 0

Take the absolute value and divide by 2: |8 - 0| / 2 = 8/2 = 4. Yes, so area is 4. Okay, that seems correct.

Then the second fill: (0,4)--(3,4)--(3,0)--cycle. So that's a triangle with vertices at (0,4), (3,4), (3,0). Let me visualize that. So from (0,4) to (3,4) is a horizontal line 3 units long, then down to (3,0), then back to (0,4). So this is a right triangle with base along the top from (0,4) to (3,4), which is 3 units, and then vertical side from (3,4) to (3,0), which is 4 units. So area would be ½ * 3 * 4 = 6.

Again, using shoelace formula to check:

Coordinates: (0,4), (3,4), (3,0), (0,4)

Sum of products:

(0*4) + (3*0) + (3*4) = 0 + 0 + 12 = 12

Minus:

(4*3) + (4*3) + (0*0) = 12 + 12 + 0 = 24

Wait, hold on, maybe I need to redo that. Wait, shoelace formula is:

For coordinates (x1,y1), (x2,y2), (x3,y3), compute:

Area = |(x1y2 + x2y3 + x3y1) - (y1x2 + y2x3 + y3x1)| / 2

So plugging in:

x1 = 0, y1 = 4

x2 = 3, y2 = 4

x3 = 3, y3 = 0

So:

First part: x1y2 + x2y3 + x3y1 = 0*4 + 3*0 + 3*4 = 0 + 0 + 12 = 12

Second part: y1x2 + y2x3 + y3x1 = 4*3 + 4*3 + 0*0 = 12 + 12 + 0 = 24

Subtract: 12 - 24 = -12, absolute value 12, divide by 2: 6. Yep, so area is 6. So that triangle has area 6.

Therefore, the total shaded area would be 4 + 6 = 10? Wait, but let me check if there is any overlap between the two shaded regions. Because if they overlap, then adding them would be double-counting. So need to check if the two triangles overlap.

First triangle: (0,4), (0,0), (2,0). Second triangle: (0,4), (3,4), (3,0). So the first triangle is on the left side, from x=0 to x=2, while the second triangle spans from x=0 to x=3. Let me think where they overlap.

The first triangle is bounded by x from 0 to 2, and the second triangle is bounded by x from 0 to 3. But they both have a common vertex at (0,4). Let me see if they share any area. The first triangle goes down to (2,0), while the second triangle goes down to (3,0). Let me think about their overlapping region.

But perhaps there is an overlap along the line from (0,4) to (something). Wait, actually, let me imagine the two triangles:

First triangle: left side, from top-left corner (0,4) down to (0,0), then right to (2,0). So this triangle is in the lower-left part of the rectangle.

Second triangle: from (0,4) to (3,4) then down to (3,0), so this triangle is the entire right side, from top to bottom, but only on the right side. Wait, but connecting (0,4) to (3,4) to (3,0) back to (0,4). So that triangle is actually a large right triangle occupying the top and right side of the rectangle.

But does the first triangle overlap with the second triangle? Let me think. The first triangle is bounded by x=0, y=0 to 4, and x=0 to 2, y=0. The second triangle is bounded by x=0 to 3, y=4 to 0, but only the area under the line from (0,4) to (3,0). Wait, actually, the second triangle is the area under the line from (0,4) to (3,0), so that line is a diagonal going from top-left to bottom-right. So the second triangle is the area under that diagonal, starting from (0,4) going to (3,0). Therefore, the first triangle is from (0,4) down to (0,0) then right to (2,0). The overlap between the two triangles would be the area that is in both triangles. So, since the first triangle is on the left side, and the second triangle is the area under the diagonal from (0,4) to (3,0). So where do they overlap?

The first triangle is under the vertical line x=2. The second triangle is under the diagonal. Let me see. The diagonal line from (0,4) to (3,0) has the equation y = (-4/3)x + 4. So at x=0, y=4; at x=3, y=0. So for x between 0 and 3, the line y = (-4/3)x + 4.

Now, the first triangle is bounded by x from 0 to 2, y from 0 up to the vertical line x=0 and the line connecting (0,0) to (2,0). Wait, no. Wait, the first triangle is formed by (0,4), (0,0), (2,0). So that triangle is bounded by the vertical line from (0,4) to (0,0), the horizontal line from (0,0) to (2,0), and the hypotenuse from (0,4) to (2,0). So that hypotenuse is a line from (0,4) to (2,0). The equation of that line is y = (-4/2)x + 4 = -2x + 4.

Therefore, the first triangle is under the line y = -2x + 4 for x from 0 to 2. The second triangle is under the line y = (-4/3)x + 4 for x from 0 to 3.

So where do these two lines intersect? Let's find the intersection point between y = -2x + 4 and y = (-4/3)x + 4.

Set -2x + 4 = (-4/3)x + 4

Subtract 4 from both sides: -2x = (-4/3)x

Add 2x to both sides: 0 = (2x - 4/3 x) = (6/3 x - 4/3 x) = (2/3)x

Therefore, 2/3 x = 0 => x = 0. Therefore, the only intersection is at x=0, y=4. So the two triangles only meet at the point (0,4) and do not overlap elsewhere. Hence, there is no overlapping area between the two shaded regions. Therefore, total shaded area is 4 + 6 = 10.

But wait, hold on. The problem says that there are twelve 1x1 squares. The total area of the rectangle is 12, so if the shaded area is 10, that would mean 10/12 = 5/6 of the area is shaded. But let me verify again. Because maybe I made a mistake in calculations.

Alternatively, maybe I should think of the figure in terms of squares.

Alternatively, maybe there's an error in the calculation. Let me think again. The first triangle: area 4. The second triangle: area 6. Total shaded area is 4 + 6 = 10. But since the entire rectangle is 3x4=12, shaded area is 10? That seems plausible, but let me check again.

Wait, but maybe the Asymptote code draws more shaded regions? Let me check the Asymptote code again:

The code says:

fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));

fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So two separate shaded regions. So if those are two triangles, then their areas are 4 and 6, as computed. Therefore total shaded area is 10. However, let me think if there's another way to calculate this.

Alternatively, maybe counting squares? Let me try to visualize the grid.

The grid is 3 units wide (x from 0 to 3) and 4 units tall (y from 0 to 4). Each square is 1x1. The shaded regions are:

First triangle: from (0,4) down to (0,0), then right to (2,0), making a triangle that covers some squares. Second triangle: from (0,4) to (3,4) down to (3,0), making another triangle.

Maybe I can count the number of squares covered by each triangle, but since triangles can partially cover squares, it's more complicated. However, if I can figure out how many squares are entirely shaded or partially shaded, but the question says "total area of the shaded region," so even if parts of squares are shaded, we need to calculate the area.

Alternatively, perhaps the Asymptote code draws two triangles which do not overlap, so total area is sum of their areas. Which would be 4 + 6 = 10. But 10 is more than half of 12. Let me check once more with coordinates.

First triangle: points (0,4), (0,0), (2,0). The area is indeed 4. Second triangle: (0,4), (3,4), (3,0). Area is 6. So total shaded area is 10. However, the problem says "twelve 1 by 1 squares form a rectangle," so the total area is 12. So 10 is less than 12. But maybe the answer is 10? Wait, but let me check with another approach.

Alternatively, maybe using coordinates. Let me try to compute the shaded area.

First triangle: The triangle with vertices at (0,4), (0,0), (2,0). Let's think about the area. Since it's a triangle, we can think of it as a right triangle with base 2 and height 4. But in reality, the base is 2 units along the x-axis and the height is 4 units along the y-axis. However, in a triangle, the area is (base * height)/2. Wait, but isn't that (base * height)/2? Wait, if we take the base as 2 units on the x-axis, then the height would be from that base to the opposite vertex, which is (0,4). The height is the vertical distance from (0,4) to the base. But the base is along the x-axis from (0,0) to (2,0). The height would be 4 units? Wait, no. Wait, if the base is on the x-axis from (0,0) to (2,0), then the height corresponding to this base is the vertical distance from this base to the opposite vertex, which is (0,4). But that vertical distance is 4 units, because the base is horizontal. Therefore, area is (base * height)/2 = (2 * 4)/2 = 4. So that's correct.

Similarly, the second triangle has base 3 units along the top (from (0,4) to (3,4)), and the height is 4 units (from (3,4) down to (3,0)), so area is (3 * 4)/2 = 6. Correct. Therefore, total shaded area is 4 + 6 = 10.

But let me check with another method. Suppose I draw the rectangle with coordinates from (0,0) to (3,4). The first triangle is from (0,4) to (0,0) to (2,0). The second triangle is from (0,4) to (3,4) to (3,0). If I imagine these two triangles, they cover different parts of the rectangle.

Alternatively, maybe there's an error with overlapping? Wait, but earlier calculation showed that they only meet at (0,4). Therefore, no overlapping.

Alternatively, perhaps the Asymptote code draws more shaded regions? Let me check again:

The Asymptote code has:

fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));

fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So two separate shaded regions. The first is the triangle from (0,4) down to (0,0) then right to (2,0). The second is the triangle from (0,4) to (3,4) down to (3,0). These are two distinct triangles. Hence, their areas add up. Therefore, total area 4 + 6 = 10. Therefore, the answer is 10? But wait, let me check if maybe the Asymptote code draws something else.

Wait, the Asymptote code also draws diagonals:

draw((0, 4)--(3, 0)); draw((0, 4)--(2, 0));

But these are just diagonals, maybe to indicate the shaded regions? Maybe the shaded regions are the triangles formed by these diagonals?

Wait, the first fill is from (0,4) to (0,0) to (2,0). So that's a triangle with vertices at (0,4), (0,0), (2,0). Then the second fill is from (0,4) to (3,4) to (3,0). So that's a triangle with vertices at (0,4), (3,4), (3,0). Therefore, the filled regions are as I thought. So total area 4 + 6 = 10. Therefore, answer is 10? But let me think again because the problem says "twelve 1 by 1 squares form a rectangle," so if the answer is 10, then it's 10 square units. But let me check once more.

Alternatively, maybe I should think in terms of the grid lines. Let me try to sketch mentally.

The entire grid is 3 columns (x=0,1,2,3) and 4 rows (y=0,1,2,3,4). Each square is 1x1. The first shaded triangle is from (0,4) to (0,0) to (2,0). So that triangle covers:

From column 0, row 4 to column 0, row 0 (the left edge), and then to column 2, row 0. So that triangle would cover parts of the squares in columns 0 and 1, rows 0 to 4.

Similarly, the second shaded triangle covers from column 0, row 4 to column 3, row 4 down to column 3, row 0. So that covers parts of columns 0 to 3, rows 0 to 4.

But since we are calculating area, not counting squares, maybe my previous calculation is correct.

But maybe the answer is 10? However, let me think if there's another way. For example, maybe the figure has overlapping lines or something else.

Wait, but according to the Asymptote code, there are two filled triangles. Let me think again if those triangles are correctly identified. For instance, maybe the first fill is not a triangle but something else?

The first fill is (0,4)--(0,0)--(2,0)--cycle. So that's three points: (0,4), (0,0), (2,0). Connecting those points makes a triangle. Similarly for the second fill. So I think the areas are correct.

Alternatively, maybe the problem is in the units? Wait, the rectangle is made of twelve 1x1 squares, so each square is 1x1, so the total area is 12. So the shaded area is 10? That seems possible. But I feel like maybe I should check again.

Alternatively, maybe I can calculate the area of the unshaded region and subtract from total area. Let's try that.

Total area is 12.

What's the unshaded area? The entire rectangle minus the two shaded triangles. But if there are no overlaps, then unshaded area is 12 - 10 = 2. But is that possible? Let's see. If you have two triangles of 4 and 6, that's 10, leaving 2. Maybe? Let me think of what's unshaded.

The first triangle (area 4) is on the left side. The second triangle (area 6) is on the top and right side. So the unshaded area would be the rest. Let me think. For example, maybe the square at (2,4) to (3,4) and (2,0) to (3,0)? Wait, not sure. Alternatively, maybe there are two small regions unshaded. But since the total area is 12, 12 - 10 = 2. If that's the case, then answer is 10. But let me check by another approach.

Alternatively, think of the entire figure. The first triangle is 4 units, second is 6 units. So total shaded area 10. Therefore, answer is 10. However, I recall that sometimes diagrams can have overlapping regions, but in this case, they don't. Hence, maybe the answer is 10. However, I need to confirm once again.

Wait, let me think of coordinates again. The first triangle covers from x=0 to x=2, y=0 to y=4. But no, it's a triangle. The area is 4. The second triangle covers from x=0 to x=3, y=0 to y=4. Its area is 6. But if you add them, you get 10. However, if you imagine the entire rectangle, 12. Maybe there's some mistake here. Wait, if I imagine the two triangles:

First triangle: area 4. Second triangle: area 6. But if you put them together, they cover parts of the rectangle. However, maybe there is an overlap in the area between the two triangles? Wait, earlier we saw that they only overlap at the point (0,4). Since that's just a single point, it doesn't contribute to area. Therefore, no overlap.

Therefore, total shaded area is indeed 4 + 6 = 10.

But to make sure, maybe I can think of specific squares.

First triangle: from (0,4) down to (0,0) then to (2,0). Let me imagine that triangle covering parts of the grid. It would cover:

In column 0, all the squares from row 0 to row 4 (since it goes from (0,4) to (0,0)), but since it's a triangle, it doesn't cover the entire column. Also, it goes from (0,0) to (2,0). So the triangle is in the lower left corner. So it would cover half of the squares along the left edge and some in the bottom.

Alternatively, maybe breaking the first triangle into smaller areas. Since it's a right triangle with base 2 and height 4, the area is 4. Similarly, the second triangle is a right triangle with base 3 and height 4, area 6. Therefore, 4 + 6 = 10. So, I think that is correct.

Therefore, I think the answer is 10. However, since the problem is from an Olympiad or competition, maybe it's 6? Wait, no. Wait, let me check again. Wait, the Asymptote code says fill((0, 4)--(0, 0)--(2, 0)--cycle... So that's a triangle that goes from (0,4) down to (0,0) then to (2,0). So this is a triangle that has a vertical side from (0,4) to (0,0) which is 4 units long, and a horizontal side from (0,0) to (2,0) which is 2 units long. Then the hypotenuse connecting (0,4) to (2,0). So the area is ½ * 2 * 4 = 4. Correct.

Then the second triangle is from (0,4) to (3,4) to (3,0). So that is a triangle with base along the top from (0,4) to (3,4) which is 3 units, and vertical side from (3,4) to (3,0) which is 4 units, so area is ½ * 3 * 4 = 6. Correct. Total area 4 + 6 = 10.

But maybe I need to check if the Asymptote code's second fill is correct. The second fill is (0, 4)--(3, 4)--(3, 0)--cycle. So that is a triangle with vertices at (0,4), (3,4), (3,0). So yes, that's the large triangle on the right side. So that seems correct.

Alternatively, maybe there's a mistake in interpreting the Asymptote code's coordinates. The Asymptote code draws a grid with vertical lines from j=0 to 3 (so x=0,1,2,3) and horizontal lines from j=0 to 4 (so y=0,1,2,3,4). Hence, the rectangle is 3 units wide and 4 units tall. So that is correct. So the total area is 3 * 4 = 12. Correct.

Therefore, I think my calculation of 10 is correct. Therefore, the answer is 10. Wait, but just to be 100% sure, let me think again. If I imagine the first triangle: from the top left corner (0,4) down to bottom left (0,0), then right to (2,0). That would cover half of the left column (since it goes from top to bottom) and a rectangle from (0,0) to (2,0). Wait, no, it's a triangle. So it would cover half of the area of a 2x4 rectangle? Wait, ½ * 2 * 4 = 4. Correct.

Similarly, the second triangle covers half of a 3x4 rectangle, which would be 6. So total shaded area is 4 + 6 = 10. Yes, that seems correct.

Therefore, I think the answer is 10. Therefore, the total area of the shaded region is 10. So \boxed{10} is the answer.

**Final Answer**
\boxed{10}

To determine the total area of the shaded region, we analyze the Asymptote diagram, which describes a 3×4 grid made up of twelve 1×1 unit squares. The shaded areas consist of two distinct triangles, as defined in the diagram.

---

### First Triangle
The first shaded triangle has vertices at the points:
- (0, 4) — the top-left corner of the rectangle
- (0, 0) — the bottom-left corner
- (2, 0) — two units to the right along the bottom edge

This triangle is a right triangle with:
- **Base** = 2 units (from (0,0) to (2,0))
- **Height** = 4 units (from (0,0) to (0,4))

Using the formula for the area of a triangle:

$$
\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4
$$

---

### Second Triangle
The second shaded triangle has vertices at:
- (0, 4) — the top-left corner
- (3, 4) — the top-right corner
- (3, 0) — the bottom-right corner

This triangle is also a right triangle with:
- **Base** = 3 units (from (0,4) to (3,4))
- **Height** = 4 units (from (3,4) to (3,0))

Again, applying the triangle area formula:

$$
\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6
$$

---

### Total Shaded Area
Since the two triangles do not overlap (they share only the point (0, 4), which has no area), their areas can be simply added:

$$
\text{Total Shaded Area} = 4 + 6 = \boxed{10}
$$

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, let's see. There's this Asymptote figure showing a rectangle made up of twelve 1 by 1 squares. The question is asking for the total area of the shaded region. Hmm. Let me try to visualize it based on the Asymptote code provided.

First, the Asymptote code draws a grid. It has vertical lines from j = 0 to 3 (so four vertical lines, which would make 3 units in width) and horizontal lines from j = 0 to 4 (five horizontal lines, so four units in height). Wait, but twelve 1x1 squares... So, if it's a rectangle, the number of squares would be width times height. Let me check. If the grid is 3 units wide (since there are 4 vertical lines, from x=0 to x=3) and 4 units tall (from y=0 to y=4), then the total area of the rectangle would be 3*4=12 squares. Yep, that matches the problem statement. So the figure is a 3x4 rectangle made up of twelve 1x1 squares.

Now, the shaded regions are filled with gray. The Asymptote code has two fill commands. Let me parse them:

First fill: (0,4)--(0,0)--(2,0)--cycle. So that's a triangle with vertices at (0,4), (0,0), and (2,0). So this is a right triangle with base along the y-axis from (0,4) to (0,0), which is 4 units, and then a base along the x-axis from (0,0) to (2,0), which is 2 units. Wait, but connecting those points would form a triangle? Wait, connecting (0,4) to (0,0) to (2,0) and back to (0,4)? So that's a right triangle with legs of length 4 and 2? Wait, no. Wait, if you connect (0,4) to (0,0) to (2,0), then back to (0,4), the triangle would have a base of 2 units (from (0,0) to (2,0)) and a height of 4 units? Wait, but in that case, the area would be (base * height)/2. But is that accurate?

Alternatively, maybe I should think in terms of coordinates. The triangle with vertices at (0,4), (0,0), and (2,0). Let me sketch this mentally. The vertical line from (0,0) to (0,4), then a horizontal line to (2,0), then back to (0,4). So the triangle is a right triangle with base along the x-axis from (0,0) to (2,0) and height along the y-axis from (0,0) to (0,4). But since it's a triangle, not a rectangle, the area would be ½ * base * height. So base is 2 units, height is 4 units? Wait, but in reality, the triangle is not a right triangle with legs 2 and 4. Wait, actually, if you have points (0,4), (0,0), (2,0), then the triangle has a vertical side of length 4, a horizontal side of length 2, and then the hypotenuse connecting (0,4) to (2,0). So yes, this is a right triangle with legs 4 and 2? Wait, no. Wait, actually, the legs would be the vertical side from (0,0) to (0,4) which is length 4, and the horizontal side from (0,0) to (2,0) which is length 2, so the area is ½ * 4 * 2 = 4. So that triangle has area 4?

But let me confirm with coordinates. Using the formula for area of a triangle with coordinates. The coordinates are (0,4), (0,0), (2,0). Using the shoelace formula:

List the coordinates:

(0,4), (0,0), (2,0), (0,4)

Compute sum of products going one way:

(0*0) + (0*0) + (2*4) = 0 + 0 + 8 = 8

Minus sum of products the other way:

(4*0) + (0*2) + (0*0) = 0 + 0 + 0 = 0

Take the absolute value and divide by 2: |8 - 0| / 2 = 8/2 = 4. Yes, so area is 4. Okay, that seems correct.

Then the second fill: (0,4)--(3,4)--(3,0)--cycle. So that's a triangle with vertices at (0,4), (3,4), (3,0). Let me visualize that. So from (0,4) to (3,4) is a horizontal line 3 units long, then down to (3,0), then back to (0,4). So this is a right triangle with base along the top from (0,4) to (3,4), which is 3 units, and then vertical side from (3,4) to (3,0), which is 4 units. So area would be ½ * 3 * 4 = 6.

Again, using shoelace formula to check:

Coordinates: (0,4), (3,4), (3,0), (0,4)

Sum of products:

(0*4) + (3*0) + (3*4) = 0 + 0 + 12 = 12

Minus:

(4*3) + (4*3) + (0*0) = 12 + 12 + 0 = 24

Wait, hold on, maybe I need to redo that. Wait, shoelace formula is:

For coordinates (x1,y1), (x2,y2), (x3,y3), compute:

Area = |(x1y2 + x2y3 + x3y1) - (y1x2 + y2x3 + y3x1)| / 2

So plugging in:

x1 = 0, y1 = 4

x2 = 3, y2 = 4

x3 = 3, y3 = 0

So:

First part: x1y2 + x2y3 + x3y1 = 0*4 + 3*0 + 3*4 = 0 + 0 + 12 = 12

Second part: y1x2 + y2x3 + y3x1 = 4*3 + 4*3 + 0*0 = 12 + 12 + 0 = 24

Subtract: 12 - 24 = -12, absolute value 12, divide by 2: 6. Yep, so area is 6. So that triangle has area 6.

Therefore, the total shaded area would be 4 + 6 = 10? Wait, but let me check if there is any overlap between the two shaded regions. Because if they overlap, then adding them would be double-counting. So need to check if the two triangles overlap.

First triangle: (0,4), (0,0), (2,0). Second triangle: (0,4), (3,4), (3,0). So the first triangle is on the left side, from x=0 to x=2, while the second triangle spans from x=0 to x=3. Let me think where they overlap.

The first triangle is bounded by x from 0 to 2, and the second triangle is bounded by x from 0 to 3. But they both have a common vertex at (0,4). Let me see if they share any area. The first triangle goes down to (2,0), while the second triangle goes down to (3,0). Let me think about their overlapping region.

But perhaps there is an overlap along the line from (0,4) to (something). Wait, actually, let me imagine the two triangles:

First triangle: left side, from top-left corner (0,4) down to (0,0), then right to (2,0). So this triangle is in the lower-left part of the rectangle.

Second triangle: from (0,4) to (3,4) then down to (3,0), so this triangle is the entire right side, from top to bottom, but only on the right side. Wait, but connecting (0,4) to (3,4) to (3,0) back to (0,4). So that triangle is actually a large right triangle occupying the top and right side of the rectangle.

But does the first triangle overlap with the second triangle? Let me think. The first triangle is bounded by x=0, y=0 to 4, and x=0 to 2, y=0. The second triangle is bounded by x=0 to 3, y=4 to 0, but only the area under the line from (0,4) to (3,0). Wait, actually, the second triangle is the area under the line from (0,4) to (3,0), so that line is a diagonal going from top-left to bottom-right. So the second triangle is the area under that diagonal, starting from (0,4) going to (3,0). Therefore, the first triangle is from (0,4) down to (0,0) then right to (2,0). The overlap between the two triangles would be the area that is in both triangles. So, since the first triangle is on the left side, and the second triangle is the area under the diagonal from (0,4) to (3,0). So where do they overlap?

The first triangle is under the vertical line x=2. The second triangle is under the diagonal. Let me see. The diagonal line from (0,4) to (3,0) has the equation y = (-4/3)x + 4. So at x=0, y=4; at x=3, y=0. So for x between 0 and 3, the line y = (-4/3)x + 4.

Now, the first triangle is bounded by x from 0 to 2, y from 0 up to the vertical line x=0 and the line connecting (0,0) to (2,0). Wait, no. Wait, the first triangle is formed by (0,4), (0,0), (2,0). So that triangle is bounded by the vertical line from (0,4) to (0,0), the horizontal line from (0,0) to (2,0), and the hypotenuse from (0,4) to (2,0). So that hypotenuse is a line from (0,4) to (2,0). The equation of that line is y = (-4/2)x + 4 = -2x + 4.

Therefore, the first triangle is under the line y = -2x + 4 for x from 0 to 2. The second triangle is under the line y = (-4/3)x + 4 for x from 0 to 3.

So where do these two lines intersect? Let's find the intersection point between y = -2x + 4 and y = (-4/3)x + 4.

Set -2x + 4 = (-4/3)x + 4

Subtract 4 from both sides: -2x = (-4/3)x

Add 2x to both sides: 0 = (2x - 4/3 x) = (6/3 x - 4/3 x) = (2/3)x

Therefore, 2/3 x = 0 => x = 0. Therefore, the only intersection is at x=0, y=4. So the two triangles only meet at the point (0,4) and do not overlap elsewhere. Hence, there is no overlapping area between the two shaded regions. Therefore, total shaded area is 4 + 6 = 10.

But wait, hold on. The problem says that there are twelve 1x1 squares. The total area of the rectangle is 12, so if the shaded area is 10, that would mean 10/12 = 5/6 of the area is shaded. But let me verify again. Because maybe I made a mistake in calculations.

Alternatively, maybe I should think of the figure in terms of squares.

Alternatively, maybe there's an error in the calculation. Let me think again. The first triangle: area 4. The second triangle: area 6. Total shaded area is 4 + 6 = 10. But since the entire rectangle is 3x4=12, shaded area is 10? That seems plausible, but let me check again.

Wait, but maybe the Asymptote code draws more shaded regions? Let me check the Asymptote code again:

The code says:

fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));

fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So two separate shaded regions. So if those are two triangles, then their areas are 4 and 6, as computed. Therefore total shaded area is 10. However, let me think if there's another way to calculate this.

Alternatively, maybe counting squares? Let me try to visualize the grid.

The grid is 3 units wide (x from 0 to 3) and 4 units tall (y from 0 to 4). Each square is 1x1. The shaded regions are:

First triangle: from (0,4) down to (0,0), then right to (2,0), making a triangle that covers some squares. Second triangle: from (0,4) to (3,4) down to (3,0), making another triangle.

Maybe I can count the number of squares covered by each triangle, but since triangles can partially cover squares, it's more complicated. However, if I can figure out how many squares are entirely shaded or partially shaded, but the question says "total area of the shaded region," so even if parts of squares are shaded, we need to calculate the area.

Alternatively, perhaps the Asymptote code draws two triangles which do not overlap, so total area is sum of their areas. Which would be 4 + 6 = 10. But 10 is more than half of 12. Let me check once more with coordinates.

First triangle: points (0,4), (0,0), (2,0). The area is indeed 4. Second triangle: (0,4), (3,4), (3,0). Area is 6. So total shaded area is 10. However, the problem says "twelve 1 by 1 squares form a rectangle," so the total area is 12. So 10 is less than 12. But maybe the answer is 10? Wait, but let me check with another approach.

Alternatively, maybe using coordinates. Let me try to compute the shaded area.

First triangle: The triangle with vertices at (0,4), (0,0), (2,0). Let's think about the area. Since it's a triangle, we can think of it as a right triangle with base 2 and height 4. But in reality, the base is 2 units along the x-axis and the height is 4 units along the y-axis. However, in a triangle, the area is (base * height)/2. Wait, but isn't that (base * height)/2? Wait, if we take the base as 2 units on the x-axis, then the height would be from that base to the opposite vertex, which is (0,4). The height is the vertical distance from (0,4) to the base. But the base is along the x-axis from (0,0) to (2,0). The height would be 4 units? Wait, no. Wait, if the base is on the x-axis from (0,0) to (2,0), then the height corresponding to this base is the vertical distance from this base to the opposite vertex, which is (0,4). But that vertical distance is 4 units, because the base is horizontal. Therefore, area is (base * height)/2 = (2 * 4)/2 = 4. So that's correct.

Similarly, the second triangle has base 3 units along the top (from (0,4) to (3,4)), and the height is 4 units (from (3,4) down to (3,0)), so area is (3 * 4)/2 = 6. Correct. Therefore, total shaded area is 4 + 6 = 10.

But let me check with another method. Suppose I draw the rectangle with coordinates from (0,0) to (3,4). The first triangle is from (0,4) to (0,0) to (2,0). The second triangle is from (0,4) to (3,4) to (3,0). If I imagine these two triangles, they cover different parts of the rectangle.

Alternatively, maybe there's an error with overlapping? Wait, but earlier calculation showed that they only meet at (0,4). Therefore, no overlapping.

Alternatively, perhaps the Asymptote code draws more shaded regions? Let me check again:

The Asymptote code has:

fill((0, 4)--(0, 0)--(2, 0)--cycle, gray(0.75));

fill((0, 4)--(3, 4)--(3, 0)--cycle, gray(0.75));

So two separate shaded regions. The first is the triangle from (0,4) down to (0,0) then right to (2,0). The second is the triangle from (0,4) to (3,4) down to (3,0). These are two distinct triangles. Hence, their areas add up. Therefore, total area 4 + 6 = 10. Therefore, the answer is 10? But wait, let me check if maybe the Asymptote code draws something else.

Wait, the Asymptote code also draws diagonals:

draw((0, 4)--(3, 0)); draw((0, 4)--(2, 0));

But these are just diagonals, maybe to indicate the shaded regions? Maybe the shaded regions are the triangles formed by these diagonals?

Wait, the first fill is from (0,4) to (0,0) to (2,0). So that's a triangle with vertices at (0,4), (0,0), (2,0). Then the second fill is from (0,4) to (3,4) to (3,0). So that's a triangle with vertices at (0,4), (3,4), (3,0). Therefore, the filled regions are as I thought. So total area 4 + 6 = 10. Therefore, answer is 10? But let me think again because the problem says "twelve 1 by 1 squares form a rectangle," so if the answer is 10, then it's 10 square units. But let me check once more.

Alternatively, maybe I should think in terms of the grid lines. Let me try to sketch mentally.

The entire grid is 3 columns (x=0,1,2,3) and 4 rows (y=0,1,2,3,4). Each square is 1x1. The first shaded triangle is from (0,4) to (0,0) to (2,0). So that triangle covers:

From column 0, row 4 to column 0, row 0 (the left edge), and then to column 2, row 0. So that triangle would cover parts of the squares in columns 0 and 1, rows 0 to 4.

Similarly, the second shaded triangle covers from column 0, row 4 to column 3, row 4 down to column 3, row 0. So that covers parts of columns 0 to 3, rows 0 to 4.

But since we are calculating area, not counting squares, maybe my previous calculation is correct.

But maybe the answer is 10? However, let me think if there's another way. For example, maybe the figure has overlapping lines or something else.

Wait, but according to the Asymptote code, there are two filled triangles. Let me think again if those triangles are correctly identified. For instance, maybe the first fill is not a triangle but something else?

The first fill is (0,4)--(0,0)--(2,0)--cycle. So that's three points: (0,4), (0,0), (2,0). Connecting those points makes a triangle. Similarly for the second fill. So I think the areas are correct.

Alternatively, maybe the problem is in the units? Wait, the rectangle is made of twelve 1x1 squares, so each square is 1x1, so the total area is 12. So the shaded area is 10? That seems possible. But I feel like maybe I should check again.

Alternatively, maybe I can calculate the area of the unshaded region and subtract from total area. Let's try that.

Total area is 12.

What's the unshaded area? The entire rectangle minus the two shaded triangles. But if there are no overlaps, then unshaded area is 12 - 10 = 2. But is that possible? Let's see. If you have two triangles of 4 and 6, that's 10, leaving 2. Maybe? Let me think of what's unshaded.

The first triangle (area 4) is on the left side. The second triangle (area 6) is on the top and right side. So the unshaded area would be the rest. Let me think. For example, maybe the square at (2,4) to (3,4) and (2,0) to (3,0)? Wait, not sure. Alternatively, maybe there are two small regions unshaded. But since the total area is 12, 12 - 10 = 2. If that's the case, then answer is 10. But let me check by another approach.

Alternatively, think of the entire figure. The first triangle is 4 units, second is 6 units. So total shaded area 10. Therefore, answer is 10. However, I recall that sometimes diagrams can have overlapping regions, but in this case, they don't. Hence, maybe the answer is 10. However, I need to confirm once again.

Wait, let me think of coordinates again. The first triangle covers from x=0 to x=2, y=0 to y=4. But no, it's a triangle. The area is 4. The second triangle covers from x=0 to x=3, y=0 to y=4. Its area is 6. But if you add them, you get 10. However, if you imagine the entire rectangle, 12. Maybe there's some mistake here. Wait, if I imagine the two triangles:

First triangle: area 4. Second triangle: area 6. But if you put them together, they cover parts of the rectangle. However, maybe there is an overlap in the area between the two triangles? Wait, earlier we saw that they only overlap at the point (0,4). Since that's just a single point, it doesn't contribute to area. Therefore, no overlap.

Therefore, total shaded area is indeed 4 + 6 = 10.

But to make sure, maybe I can think of specific squares.

First triangle: from (0,4) down to (0,0) then to (2,0). Let me imagine that triangle covering parts of the grid. It would cover:

In column 0, all the squares from row 0 to row 4 (since it goes from (0,4) to (0,0)), but since it's a triangle, it doesn't cover the entire column. Also, it goes from (0,0) to (2,0). So the triangle is in the lower left corner. So it would cover half of the squares along the left edge and some in the bottom.

Alternatively, maybe breaking the first triangle into smaller areas. Since it's a right triangle with base 2 and height 4, the area is 4. Similarly, the second triangle is a right triangle with base 3 and height 4, area 6. Therefore, 4 + 6 = 10. So, I think that is correct.

Therefore, I think the answer is 10. However, since the problem is from an Olympiad or competition, maybe it's 6? Wait, no. Wait, let me check again. Wait, the Asymptote code says fill((0, 4)--(0, 0)--(2, 0)--cycle... So that's a triangle that goes from (0,4) down to (0,0) then to (2,0). So this is a triangle that has a vertical side from (0,4) to (0,0) which is 4 units long, and a horizontal side from (0,0) to (2,0) which is 2 units long. Then the hypotenuse connecting (0,4) to (2,0). So the area is ½ * 2 * 4 = 4. Correct.

Then the second triangle is from (0,4) to (3,4) to (3,0). So that is a triangle with base along the top from (0,4) to (3,4) which is 3 units, and vertical side from (3,4) to (3,0) which is 4 units, so area is ½ * 3 * 4 = 6. Correct. Total area 4 + 6 = 10.

But maybe I need to check if the Asymptote code's second fill is correct. The second fill is (0, 4)--(3, 4)--(3, 0)--cycle. So that is a triangle with vertices at (0,4), (3,4), (3,0). So yes, that's the large triangle on the right side. So that seems correct.

Alternatively, maybe there's a mistake in interpreting the Asymptote code's coordinates. The Asymptote code draws a grid with vertical lines from j=0 to 3 (so x=0,1,2,3) and horizontal lines from j=0 to 4 (so y=0,1,2,3,4). Hence, the rectangle is 3 units wide and 4 units tall. So that is correct. So the total area is 3 * 4 = 12. Correct.

Therefore, I think my calculation of 10 is correct. Therefore, the answer is 10. Wait, but just to be 100% sure, let me think again. If I imagine the first triangle: from the top left corner (0,4) down to bottom left (0,0), then right to (2,0). That would cover half of the left column (since it goes from top to bottom) and a rectangle from (0,0) to (2,0). Wait, no, it's a triangle. So it would cover half of the area of a 2x4 rectangle? Wait, ½ * 2 * 4 = 4. Correct.

Similarly, the second triangle covers half of a 3x4 rectangle, which would be 6. So total shaded area is 4 + 6 = 10. Yes, that seems correct.

Therefore, I think the answer is 10. Therefore, the total area of the shaded region is 10. So \boxed{10} is the answer.

**Final Answer**
\boxed{10}

To determine the total area of the shaded region, we analyze the Asymptote diagram, which describes a 3×4 grid made up of twelve 1×1 unit squares. The shaded areas consist of two distinct triangles, as defined in the diagram.

---

### First Triangle
The first shaded triangle has vertices at the points:
- (0, 4) — the top-left corner of the rectangle
- (0, 0) — the bottom-left corner
- (2, 0) — two units to the right along the bottom edge

This triangle is a right triangle with:
- **Base** = 2 units (from (0,0) to (2,0))
- **Height** = 4 units (from (0,0) to (0,4))

Using the formula for the area of a triangle:

$$
\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4
$$

---

### Second Triangle
The second shaded triangle has vertices at:
- (0, 4) — the top-left corner
- (3, 4) — the top-right corner
- (3, 0) — the bottom-right corner

This triangle is also a right triangle with:
- **Base** = 3 units (from (0,4) to (3,4))
- **Height** = 4 units (from (3,4) to (3,0))

Again, applying the triangle area formula:

$$
\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6
$$

---

### Total Shaded Area
Since the two triangles do not overlap (they share only the point (0, 4), which has no area), their areas can be simply added:

$$
\text{Total Shaded Area} = 4 + 6 = \boxed{10}
$$

---

### Final Answer

$$
\boxed{10}
$$

Solution:
The total area of the rectangle is $3 \times 4 =12$.

The total area of the shaded regions equals the total area of the rectangle (12) minus the area of the unshaded region.

The unshaded region is a triangle with base of length 1 and height 4; the area of this region is $\frac{1}{2}(1)(4)=2$.

Therefore, the total area of the shaded regions is $12 - 2 = \boxed{10}$.


Explanation:
Okay, so I need to compute the infinite series $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$, where $F_n$ is the Fibonacci sequence with $F_0 = 0$ and $F_1 = 1$. Hmm, Fibonacci numbers... I remember they follow the recurrence relation $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 2$, right? Let me write that down:

$F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$, with $F_0 = 0$ and $F_1 = 1$.

So the problem is about summing terms of the form $\frac{F_{n+1}}{F_n F_{n+2}}$ from $n=1$ to infinity. Let me think about how to approach this.

First, maybe I can try writing out the first few terms to see if there's a pattern or telescoping behavior. Let me compute the first few terms manually.

Let me recall the Fibonacci numbers:

- $F_0 = 0$
- $F_1 = 1$
- $F_2 = 1$
- $F_3 = 2$
- $F_4 = 3$
- $F_5 = 5$
- $F_6 = 8$
- $F_7 = 13$
- $F_8 = 21$
- $F_9 = 34$
- $F_{10} = 55$
- ... and so on.

So let me compute the first few terms of the series:

For $n=1$: $\frac{F_2}{F_1 F_3} = \frac{1}{1 \times 2} = 1/2$

For $n=2$: $\frac{F_3}{F_2 F_4} = \frac{2}{1 \times 3} = 2/3$

For $n=3$: $\frac{F_4}{F_3 F_5} = \frac{3}{2 \times 5} = 3/10$

For $n=4$: $\frac{F_5}{F_4 F_6} = \frac{5}{3 \times 8} = 5/24$

For $n=5$: $\frac{F_6}{F_5 F_7} = \frac{8}{5 \times 13} = 8/65$

So the terms are 1/2, 2/3, 3/10, 5/24, 8/65, etc. Let me add them up partially to see if there's a pattern:

First term: 1/2 ≈ 0.5

First two terms: 1/2 + 2/3 = 7/6 ≈ 1.1667

First three terms: 7/6 + 3/10 = (35/30 + 9/30) = 44/30 ≈ 1.4667

First four terms: 44/30 + 5/24 = (44/30 + 5/24) = Let's convert to common denominator. 30 and 24 have LCM of 120. So 44/30 = 176/120, 5/24 = 25/120. Sum is 201/120 ≈ 1.675.

First five terms: 201/120 + 8/65. Let me compute this. Convert to common denominator of 120 and 65. LCM of 120 and 65 is 1560? Let me check:

65 factors into 5×13, 120 into 2^3×3×5. So LCM is 2^3×3×5×13 = 8×3×5×13 = 24×65 = 1560. So 201/120 = (201×13)/1560 = 2613/1560. 8/65 = (8×24)/1560 = 192/1560. Sum is 2613 + 192 = 2805 / 1560 ≈ 1.798.

Hmm, so partial sums are approaching something... maybe approaching 2? But let's check more terms? But maybe there's a telescoping series here?

Alternatively, maybe I can find a telescoping expression for the general term $\frac{F_{n+1}}{F_n F_{n+2}}$.

Telescoping series often involve expressing terms as differences of consecutive terms in another sequence. So perhaps I can manipulate the term into something like $a_n - a_{n+1}$ or similar.

Let me think about Fibonacci identities. I remember that there are several identities involving Fibonacci numbers. For example, the identity $F_{n+2} = F_{n+1} + F_n$. Maybe I can use that here.

Let me write the denominator $F_n F_{n+2}$ as $F_n (F_{n+1} + F_n)$. Let me try substituting that:

So $\frac{F_{n+1}}{F_n (F_{n+1} + F_n)}$.

Hmm, that might be helpful. Let me see if I can manipulate this expression.

Alternatively, maybe I can relate this to reciprocals or differences.

Alternatively, perhaps express $\frac{F_{n+1}}{F_n F_{n+2}}$ as something involving reciprocals of Fibonacci numbers. Let me think.

Alternatively, since Fibonacci numbers satisfy certain recurrence relations, maybe we can express the term as a difference.

Let me try to think of telescoping terms. Suppose I can find constants $a$ and $b$ such that:

$\frac{F_{n+1}}{F_n F_{n+2}} = \frac{a}{F_n} - \frac{a}{F_{n+1}}$ or something similar.

Wait, let me try to manipulate the term.

First, note that $F_{n+2} = F_{n+1} + F_n$. Therefore, we can write:

$\frac{F_{n+1}}{F_n F_{n+2}} = \frac{F_{n+1}}{F_n (F_{n+1} + F_n)}$

Let me divide numerator and denominator by $F_{n+1}$:

$\frac{1}{F_n (1 + \frac{F_n}{F_{n+1}})}$

Hmm, not sure if helpful. Alternatively, maybe split the fraction:

Let me consider:

Let me denote $x = F_{n+1}$, $y = F_n$. Then $F_{n+2} = x + y$. Then the term is $x/(y(x + y))$.

So $x/(y(x + y)) = [something]$. Maybe we can split this fraction into partial fractions?

Let me try to write:

Let me suppose that $x/(y(x + y)) = A/y + B/(x + y)$. Let me solve for A and B:

Multiply both sides by $y(x + y)$:

x = A(x + y) + B y

Expanding: x = A x + A y + B y = A x + (A + B)y

Therefore, equating coefficients:

For x: 1 = A

For constant term (in y): 0 = A + B => B = -A = -1

Therefore,

x/(y(x + y)) = 1/y - 1/(x + y)

Therefore, substituting back x = F_{n+1}, y = F_n:

$\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}$

Wait, that seems promising!

Let me check this with n=1:

For n=1: left side is $\frac{F_2}{F_1 F_3} = 1/(1*2) = 1/2$. Right side: 1/F_1 - 1/F_3 = 1/1 - 1/2 = 1 - 1/2 = 1/2. Correct.

For n=2: left side is 2/(1*3) = 2/3. Right side: 1/F_2 - 1/F_4 = 1/1 - 1/3 = 2/3. Correct.

For n=3: left side is 3/(2*5) = 3/10. Right side: 1/F_3 - 1/F_5 = 1/2 - 1/5 = 3/10. Correct. Perfect! So the identity holds.

Therefore, the term $\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}$.

Therefore, our original series becomes:

$\sum_{n=1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+2}} \right )$.

This is a telescoping series! Because when we expand the terms, most terms will cancel out.

Let me write out the first few terms:

For n=1: 1/F_1 - 1/F_3

For n=2: 1/F_2 - 1/F_4

For n=3: 1/F_3 - 1/F_5

For n=4: 1/F_4 - 1/F_6

For n=5: 1/F_5 - 1/F_7

...

So when we sum from n=1 to N, the series becomes:

(1/F_1 - 1/F_3) + (1/F_2 - 1/F_4) + (1/F_3 - 1/F_5) + (1/F_4 - 1/F_6) + ... + (1/F_N - 1/F_{N+2} )

Now, let's see cancellation. The -1/F_3 cancels with the +1/F_3 from the next term. Similarly, -1/F_4 cancels with +1/F_4 from the next term, etc. So all intermediate terms cancel, and we are left with the first few positive terms and the last few negative terms.

Specifically, the partial sum up to N is:

1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}

Because all the intermediate terms from -1/F_3 to -1/F_N cancel with the subsequent +1/F_3 to +1/F_N.

So the partial sum S_N = [1/F_1 + 1/F_2] - [1/F_{N+1} + 1/F_{N+2}]

Therefore, taking the limit as N approaches infinity:

Sum from n=1 to infinity = lim_{N→∞} [1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}]

We know that F_1 = 1 and F_2 = 1, so 1/F_1 + 1/F_2 = 1 + 1 = 2.

Now, we need to compute the limit of [ -1/F_{N+1} - 1/F_{N+2} ] as N approaches infinity.

But Fibonacci numbers grow exponentially (since they follow the golden ratio), so 1/F_{N+1} and 1/F_{N+2} approach zero as N approaches infinity. Therefore, the limit becomes:

2 - 0 - 0 = 2.

Therefore, the sum converges to 2.

Wait, but let me check with the partial sums I calculated earlier. For example, when I added up the first five terms, I had approximately 1.798, which is less than 2. So as more terms are added, it approaches 2. That seems plausible.

But let me check for N=1: partial sum is 1/F_1 -1/F_3 = 1 - 1/2 = 1/2, which matches the first term. For N=2: 1/F_1 +1/F_2 -1/F_3 -1/F_4 = 1 + 1 - 1/2 -1/3 = 2 - 5/6 = 7/6 ≈ 1.166..., which matches the first two terms. For N=3: 2 - 1/F_4 - 1/F_5 = 2 - 1/3 - 1/5 = 2 - 8/15 = 22/15 ≈ 1.466..., which matches the first three terms. So yes, seems correct.

So the telescoping works, and the sum converges to 2. Therefore, the answer should be 2.

But let me double-check my steps again.

First, I used the identity $\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}$. Verified with n=1,2,3, and seems correct. Then the telescoping sum gives the partial sum as 2 - [1/F_{N+1} + 1/F_{N+2}], which tends to 2. So the conclusion is 2. Seems solid.

But just to be thorough, let me check another term. For n=4, term is 5/(3*8) = 5/24 ≈ 0.2083. The partial sum up to n=4 would be 2 - [1/F_5 + 1/F_6] = 2 - [1/5 + 1/8] = 2 - 13/40 = 67/40 = 1.675, which matches my earlier calculation (201/120 is 1.675). Yes. So that's correct.

Therefore, I think this is correct. Therefore, the answer is 2.

**Final Answer**
\boxed{2}

To evaluate the infinite series
$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},
$$
we begin by analyzing the structure of the general term and the recurrence relation satisfied by the Fibonacci numbers.

---

### Step 1: Fibonacci Recurrence
The Fibonacci sequence is defined as:
$$
F_0 = 0, \quad F_1 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n \geq 2.
$$
This recurrence will be useful in rewriting the general term of the series.

---

### Step 2: Simplify the General Term
We are summing:
$$
\frac{F_{n + 1}}{F_n F_{n + 2}}.
$$
Notice that $F_{n + 2} = F_{n + 1} + F_n$ by the Fibonacci recurrence. Substituting:
$$
\frac{F_{n+1}}{F_n (F_{n+1} + F_n)}.
$$
Let’s apply partial fraction decomposition here. Let’s write:
$$
\frac{F_{n+1}}{F_n (F_{n+1} + F_n)} = \frac{1}{F_n} - \frac{1}{F_{n+2}}.
$$
This is a known identity for Fibonacci numbers, and we can verify it by simplifying the right-hand side:
$$
\frac{1}{F_n} - \frac{1}{F_{n+2}} = \frac{F_{n+2} - F_n}{F_n F_{n+2}} = \frac{(F_{n+1} + F_n) - F_n}{F_n F_{n+2}} = \frac{F_{n+1}}{F_n F_{n+2}},
$$
which matches the left-hand side. So the decomposition is correct.

---

### Step 3: Rewrite the Series
Substituting back into the series, we have:
$$
\sum_{n=1}^\infty \frac{F_{n+1}}{F_n F_{n+2}} = \sum_{n=1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+2}} \right).
$$
This is a **telescoping series**. Let's write out the first few terms:
$$
\left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \left( \frac{1}{F_4} - \frac{1}{F_6} \right) + \cdots
$$
All the intermediate terms cancel:
$$
\left( \frac{1}{F_1} + \frac{1}{F_2} \right) - \lim_{N \to \infty} \left( \frac{1}{F_{N+1}} + \frac{1}{F_{N+2}} \right).
$$

---

### Step 4: Evaluate the Limit
We know:
- $F_1 = 1$, $F_2 = 1$, so the constant terms give:
$$
\frac{1}{F_1} + \frac{1}{F_2} = 1 + 1 = 2.
$$
- As $n \to \infty$, Fibonacci numbers grow exponentially. Thus:
$$
\lim_{N \to \infty} \left( \frac{1}{F_{N+1}} + \frac{1}{F_{N+2}} \right) = 0.
$$

Therefore, the sum becomes:
$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = 2 - 0 = 2.
$$

---

### Final Answer
$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to compute the infinite series $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$, where $F_n$ is the Fibonacci sequence with $F_0 = 0$ and $F_1 = 1$. Hmm, Fibonacci numbers... I remember they follow the recurrence relation $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 2$, right? Let me write that down:

$F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$, with $F_0 = 0$ and $F_1 = 1$.

So the problem is about summing terms of the form $\frac{F_{n+1}}{F_n F_{n+2}}$ from $n=1$ to infinity. Let me think about how to approach this.

First, maybe I can try writing out the first few terms to see if there's a pattern or telescoping behavior. Let me compute the first few terms manually.

Let me recall the Fibonacci numbers:

- $F_0 = 0$
- $F_1 = 1$
- $F_2 = 1$
- $F_3 = 2$
- $F_4 = 3$
- $F_5 = 5$
- $F_6 = 8$
- $F_7 = 13$
- $F_8 = 21$
- $F_9 = 34$
- $F_{10} = 55$
- ... and so on.

So let me compute the first few terms of the series:

For $n=1$: $\frac{F_2}{F_1 F_3} = \frac{1}{1 \times 2} = 1/2$

For $n=2$: $\frac{F_3}{F_2 F_4} = \frac{2}{1 \times 3} = 2/3$

For $n=3$: $\frac{F_4}{F_3 F_5} = \frac{3}{2 \times 5} = 3/10$

For $n=4$: $\frac{F_5}{F_4 F_6} = \frac{5}{3 \times 8} = 5/24$

For $n=5$: $\frac{F_6}{F_5 F_7} = \frac{8}{5 \times 13} = 8/65$

So the terms are 1/2, 2/3, 3/10, 5/24, 8/65, etc. Let me add them up partially to see if there's a pattern:

First term: 1/2 ≈ 0.5

First two terms: 1/2 + 2/3 = 7/6 ≈ 1.1667

First three terms: 7/6 + 3/10 = (35/30 + 9/30) = 44/30 ≈ 1.4667

First four terms: 44/30 + 5/24 = (44/30 + 5/24) = Let's convert to common denominator. 30 and 24 have LCM of 120. So 44/30 = 176/120, 5/24 = 25/120. Sum is 201/120 ≈ 1.675.

First five terms: 201/120 + 8/65. Let me compute this. Convert to common denominator of 120 and 65. LCM of 120 and 65 is 1560? Let me check:

65 factors into 5×13, 120 into 2^3×3×5. So LCM is 2^3×3×5×13 = 8×3×5×13 = 24×65 = 1560. So 201/120 = (201×13)/1560 = 2613/1560. 8/65 = (8×24)/1560 = 192/1560. Sum is 2613 + 192 = 2805 / 1560 ≈ 1.798.

Hmm, so partial sums are approaching something... maybe approaching 2? But let's check more terms? But maybe there's a telescoping series here?

Alternatively, maybe I can find a telescoping expression for the general term $\frac{F_{n+1}}{F_n F_{n+2}}$.

Telescoping series often involve expressing terms as differences of consecutive terms in another sequence. So perhaps I can manipulate the term into something like $a_n - a_{n+1}$ or similar.

Let me think about Fibonacci identities. I remember that there are several identities involving Fibonacci numbers. For example, the identity $F_{n+2} = F_{n+1} + F_n$. Maybe I can use that here.

Let me write the denominator $F_n F_{n+2}$ as $F_n (F_{n+1} + F_n)$. Let me try substituting that:

So $\frac{F_{n+1}}{F_n (F_{n+1} + F_n)}$.

Hmm, that might be helpful. Let me see if I can manipulate this expression.

Alternatively, maybe I can relate this to reciprocals or differences.

Alternatively, perhaps express $\frac{F_{n+1}}{F_n F_{n+2}}$ as something involving reciprocals of Fibonacci numbers. Let me think.

Alternatively, since Fibonacci numbers satisfy certain recurrence relations, maybe we can express the term as a difference.

Let me try to think of telescoping terms. Suppose I can find constants $a$ and $b$ such that:

$\frac{F_{n+1}}{F_n F_{n+2}} = \frac{a}{F_n} - \frac{a}{F_{n+1}}$ or something similar.

Wait, let me try to manipulate the term.

First, note that $F_{n+2} = F_{n+1} + F_n$. Therefore, we can write:

$\frac{F_{n+1}}{F_n F_{n+2}} = \frac{F_{n+1}}{F_n (F_{n+1} + F_n)}$

Let me divide numerator and denominator by $F_{n+1}$:

$\frac{1}{F_n (1 + \frac{F_n}{F_{n+1}})}$

Hmm, not sure if helpful. Alternatively, maybe split the fraction:

Let me consider:

Let me denote $x = F_{n+1}$, $y = F_n$. Then $F_{n+2} = x + y$. Then the term is $x/(y(x + y))$.

So $x/(y(x + y)) = [something]$. Maybe we can split this fraction into partial fractions?

Let me try to write:

Let me suppose that $x/(y(x + y)) = A/y + B/(x + y)$. Let me solve for A and B:

Multiply both sides by $y(x + y)$:

x = A(x + y) + B y

Expanding: x = A x + A y + B y = A x + (A + B)y

Therefore, equating coefficients:

For x: 1 = A

For constant term (in y): 0 = A + B => B = -A = -1

Therefore,

x/(y(x + y)) = 1/y - 1/(x + y)

Therefore, substituting back x = F_{n+1}, y = F_n:

$\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}$

Wait, that seems promising!

Let me check this with n=1:

For n=1: left side is $\frac{F_2}{F_1 F_3} = 1/(1*2) = 1/2$. Right side: 1/F_1 - 1/F_3 = 1/1 - 1/2 = 1 - 1/2 = 1/2. Correct.

For n=2: left side is 2/(1*3) = 2/3. Right side: 1/F_2 - 1/F_4 = 1/1 - 1/3 = 2/3. Correct.

For n=3: left side is 3/(2*5) = 3/10. Right side: 1/F_3 - 1/F_5 = 1/2 - 1/5 = 3/10. Correct. Perfect! So the identity holds.

Therefore, the term $\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}$.

Therefore, our original series becomes:

$\sum_{n=1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+2}} \right )$.

This is a telescoping series! Because when we expand the terms, most terms will cancel out.

Let me write out the first few terms:

For n=1: 1/F_1 - 1/F_3

For n=2: 1/F_2 - 1/F_4

For n=3: 1/F_3 - 1/F_5

For n=4: 1/F_4 - 1/F_6

For n=5: 1/F_5 - 1/F_7

...

So when we sum from n=1 to N, the series becomes:

(1/F_1 - 1/F_3) + (1/F_2 - 1/F_4) + (1/F_3 - 1/F_5) + (1/F_4 - 1/F_6) + ... + (1/F_N - 1/F_{N+2} )

Now, let's see cancellation. The -1/F_3 cancels with the +1/F_3 from the next term. Similarly, -1/F_4 cancels with +1/F_4 from the next term, etc. So all intermediate terms cancel, and we are left with the first few positive terms and the last few negative terms.

Specifically, the partial sum up to N is:

1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}

Because all the intermediate terms from -1/F_3 to -1/F_N cancel with the subsequent +1/F_3 to +1/F_N.

So the partial sum S_N = [1/F_1 + 1/F_2] - [1/F_{N+1} + 1/F_{N+2}]

Therefore, taking the limit as N approaches infinity:

Sum from n=1 to infinity = lim_{N→∞} [1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}]

We know that F_1 = 1 and F_2 = 1, so 1/F_1 + 1/F_2 = 1 + 1 = 2.

Now, we need to compute the limit of [ -1/F_{N+1} - 1/F_{N+2} ] as N approaches infinity.

But Fibonacci numbers grow exponentially (since they follow the golden ratio), so 1/F_{N+1} and 1/F_{N+2} approach zero as N approaches infinity. Therefore, the limit becomes:

2 - 0 - 0 = 2.

Therefore, the sum converges to 2.

Wait, but let me check with the partial sums I calculated earlier. For example, when I added up the first five terms, I had approximately 1.798, which is less than 2. So as more terms are added, it approaches 2. That seems plausible.

But let me check for N=1: partial sum is 1/F_1 -1/F_3 = 1 - 1/2 = 1/2, which matches the first term. For N=2: 1/F_1 +1/F_2 -1/F_3 -1/F_4 = 1 + 1 - 1/2 -1/3 = 2 - 5/6 = 7/6 ≈ 1.166..., which matches the first two terms. For N=3: 2 - 1/F_4 - 1/F_5 = 2 - 1/3 - 1/5 = 2 - 8/15 = 22/15 ≈ 1.466..., which matches the first three terms. So yes, seems correct.

So the telescoping works, and the sum converges to 2. Therefore, the answer should be 2.

But let me double-check my steps again.

First, I used the identity $\frac{F_{n+1}}{F_n F_{n+2}} = \frac{1}{F_n} - \frac{1}{F_{n+2}}$. Verified with n=1,2,3, and seems correct. Then the telescoping sum gives the partial sum as 2 - [1/F_{N+1} + 1/F_{N+2}], which tends to 2. So the conclusion is 2. Seems solid.

But just to be thorough, let me check another term. For n=4, term is 5/(3*8) = 5/24 ≈ 0.2083. The partial sum up to n=4 would be 2 - [1/F_5 + 1/F_6] = 2 - [1/5 + 1/8] = 2 - 13/40 = 67/40 = 1.675, which matches my earlier calculation (201/120 is 1.675). Yes. So that's correct.

Therefore, I think this is correct. Therefore, the answer is 2.

**Final Answer**
\boxed{2}

To evaluate the infinite series
$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},
$$
we begin by analyzing the structure of the general term and the recurrence relation satisfied by the Fibonacci numbers.

---

### Step 1: Fibonacci Recurrence
The Fibonacci sequence is defined as:
$$
F_0 = 0, \quad F_1 = 1, \quad F_n = F_{n-1} + F_{n-2} \text{ for } n \geq 2.
$$
This recurrence will be useful in rewriting the general term of the series.

---

### Step 2: Simplify the General Term
We are summing:
$$
\frac{F_{n + 1}}{F_n F_{n + 2}}.
$$
Notice that $F_{n + 2} = F_{n + 1} + F_n$ by the Fibonacci recurrence. Substituting:
$$
\frac{F_{n+1}}{F_n (F_{n+1} + F_n)}.
$$
Let’s apply partial fraction decomposition here. Let’s write:
$$
\frac{F_{n+1}}{F_n (F_{n+1} + F_n)} = \frac{1}{F_n} - \frac{1}{F_{n+2}}.
$$
This is a known identity for Fibonacci numbers, and we can verify it by simplifying the right-hand side:
$$
\frac{1}{F_n} - \frac{1}{F_{n+2}} = \frac{F_{n+2} - F_n}{F_n F_{n+2}} = \frac{(F_{n+1} + F_n) - F_n}{F_n F_{n+2}} = \frac{F_{n+1}}{F_n F_{n+2}},
$$
which matches the left-hand side. So the decomposition is correct.

---

### Step 3: Rewrite the Series
Substituting back into the series, we have:
$$
\sum_{n=1}^\infty \frac{F_{n+1}}{F_n F_{n+2}} = \sum_{n=1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n+2}} \right).
$$
This is a **telescoping series**. Let's write out the first few terms:
$$
\left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \left( \frac{1}{F_4} - \frac{1}{F_6} \right) + \cdots
$$
All the intermediate terms cancel:
$$
\left( \frac{1}{F_1} + \frac{1}{F_2} \right) - \lim_{N \to \infty} \left( \frac{1}{F_{N+1}} + \frac{1}{F_{N+2}} \right).
$$

---

### Step 4: Evaluate the Limit
We know:
- $F_1 = 1$, $F_2 = 1$, so the constant terms give:
$$
\frac{1}{F_1} + \frac{1}{F_2} = 1 + 1 = 2.
$$
- As $n \to \infty$, Fibonacci numbers grow exponentially. Thus:
$$
\lim_{N \to \infty} \left( \frac{1}{F_{N+1}} + \frac{1}{F_{N+2}} \right) = 0.
$$

Therefore, the sum becomes:
$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = 2 - 0 = 2.
$$

---

### Final Answer
$$
\boxed{2}
$$

Solution:
Since $F_{n + 1} = F_{n + 2} - F_n,$
\[\frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{F_{n + 2} - F_n}{F_n F_{n + 2}} = \frac{1}{F_n} - \frac{1}{F_{n + 2}}.\]Then
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{F_n F_{n + 2}} &= \left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \dotsb \\
&= \frac{1}{F_1} + \frac{1}{F_2} \\
&= \boxed{2}.
\end{align*}