Explanation:
Okay, so I need to compute the infinite sum $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$, 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 $n \geq 2$, with $F_0 = 0$ and $F_1 = 1$.

First, maybe I should write out the first few terms of the sequence to get a sense of what's going on. 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 the sum starts at $n = 1$, so let's compute the first few terms of the sum:

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

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

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

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

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

So adding these up: 1/2 + 2/3 + 3/10 + 5/24 + 8/65 + ...

Hmm, not sure if that helps directly. Maybe I need to find a telescoping series or some identity involving Fibonacci numbers.

I remember that Fibonacci numbers have some identities. For example, Cassini's identity: $F_{n+1}F_{n-1} - F_n^2 = (-1)^n$. Maybe that's useful? Or perhaps some other identity?

Alternatively, maybe I can express the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$ in terms of differences of reciprocals or something similar. Let me think.

First, let's recall the Fibonacci recurrence: $F_{n + 2} = F_{n + 1} + F_n$. Maybe I can use this to rewrite the denominator. Let me try:

Denominator is $F_n F_{n + 2} = F_n (F_{n + 1} + F_n) = F_n F_{n + 1} + F_n^2$. Not sure if that helps yet.

Alternatively, maybe express the term as something involving $F_{n + 1}/F_n$ or $F_{n + 1}/F_{n + 2}$. Let me think about the ratio of consecutive Fibonacci numbers.

I know that $F_{n + 1}/F_n$ approaches the golden ratio $\phi = (1 + \sqrt{5})/2$ as $n$ becomes large, but I don't know if that helps here.

Alternatively, maybe I can manipulate the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$ using the recurrence relation. Let me try to express it in terms of differences.

Let me consider the identity $F_{n + 2} = F_{n + 1} + F_n$. Let me solve for $F_{n + 1}$: $F_{n + 1} = F_{n + 2} - F_n$. Wait, but substituting that into the numerator:

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

Oh! That seems promising! So the term simplifies to $\frac{1}{F_n} - \frac{1}{F_{n + 2}}$. Let me check that with an example. Let's take n=1:

Original term: $F_2/(F_1 F_3) = 1/(1*2) = 1/2$.

According to the expression $\frac{1}{F_1} - \frac{1}{F_3} = 1/1 - 1/2 = 1 - 1/2 = 1/2. Correct.

Another example, n=2: original term is 2/(1*3) = 2/3. The expression is 1/F_2 - 1/F_4 = 1/1 - 1/3 = 2/3. Correct again.

n=3: original term is 3/(2*5) = 3/10. The expression is 1/F_3 - 1/F_5 = 1/2 - 1/5 = 5/10 - 2/10 = 3/10. Correct.

Great! So the general term simplifies to $\frac{1}{F_n} - \frac{1}{F_{n + 2}}$. Therefore, the sum becomes:

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

This looks like a telescoping series! Let me write out the first few terms to see:

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

n=2: 1/F_2 - 1/F_4

n=3: 1/F_3 - 1/F_5

n=4: 1/F_4 - 1/F_6

n=5: 1/F_5 - 1/F_7

...

So when we add them up, most terms should cancel. Let's see:

Sum = (1/F_1 - 1/F_3) + (1/F_2 - 1/F_4) + (1/F_3 - 1/F_5) + (1/F_4 - 1/F_6) + ...

Indeed, the -1/F_3 cancels with +1/F_3 from the next term, similarly -1/F_4 cancels with +1/F_4, etc. So all the intermediate terms cancel, and we are left with the first few terms that don't get canceled.

Specifically, the sum telescopes to:

1/F_1 + 1/F_2 - lim_{k→∞} [1/F_{k + 1} + 1/F_{k + 2}]

Wait, let me think again. Let me write the partial sum up to N terms:

S_N = sum_{n=1}^N [1/F_n - 1/F_{n + 2}] = [1/F_1 - 1/F_3] + [1/F_2 - 1/F_4] + [1/F_3 - 1/F_5] + ... + [1/F_N - 1/F_{N + 2}]

Now, grouping the terms:

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

Wait, actually, the cancellation is between -1/F_{n + 2} and the next term's +1/F_{n + 2}? Wait, let me check:

Wait, for each term, the negative part is -1/F_{n + 2}, and the next term's positive part is +1/F_{n + 1}... Wait, no. Let me think again.

Wait, the first term is 1/F_1 - 1/F_3

Second term is 1/F_2 - 1/F_4

Third term is 1/F_3 - 1/F_5

Fourth term is 1/F_4 - 1/F_6

...

So when you add them up, the -1/F_3 cancels with the +1/F_3 in the third term. Similarly, -1/F_4 cancels with +1/F_4 in the fourth term, etc. So the partial sum S_N is:

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

Because all the intermediate terms cancel. Let me check with N=2:

S_2 = [1/F_1 - 1/F_3] + [1/F_2 - 1/F_4] = 1/F_1 + 1/F_2 - 1/F_3 - 1/F_4. Which is 1/F_1 + 1/F_2 - 1/F_{2 + 1} - 1/F_{2 + 2} = yes. Similarly for N=3: S_3 = 1/F_1 + 1/F_2 - 1/F_4 - 1/F_5. Which is 1/F_1 + 1/F_2 - 1/F_{3 + 1} - 1/F_{3 + 2}. So yes, general formula is:

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 = lim_{N→∞} [1/F_1 + 1/F_2 - 1/F_{N + 1} - 1/F_{N + 2}]

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

Now, we need to evaluate the limit of -1/F_{N + 1} - 1/F_{N + 2} as N approaches infinity. Since Fibonacci numbers grow exponentially (they grow like φ^n / sqrt(5)), so 1/F_{N + 1} and 1/F_{N + 2} both approach 0 as N approaches infinity. Therefore, the limit is 0 - 0 = 0.

Therefore, the total sum is 2 - 0 = 2. Wait, so the sum equals 2? Let me check with the partial sums I calculated earlier.

Earlier, the first term was 1/2, then adding 2/3 gives 1/2 + 2/3 = 7/6 ≈ 1.166..., then adding 3/10 gives 7/6 + 3/10 = 35/30 + 9/30 = 44/30 ≈ 1.466..., then adding 5/24 gives 44/30 + 5/24 = (176 + 25)/120 = 201/120 ≈ 1.675, then adding 8/65 gives 201/120 + 8/65 = let's compute that:

Convert to common denominator, which is 120*65 = 7800. 201/120 = (201*65)/7800 = let me compute 201*65:

200*65 = 13,000; 1*65 = 65; total 13,065. Then 8/65 = (8*120)/7800 = 960/7800. So total is 13,065 + 960 = 14,025 / 7800 ≈ 1.8.

So after 5 terms, the partial sum is about 1.8, which is approaching 2. So that seems plausible.

Therefore, the sum converges to 2.

Wait, but let me check again the telescoping sum. The partial sum S_N is 1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2} = 1 + 1 - [1/F_{N+1} + 1/F_{N+2}]. So as N approaches infinity, the subtracted terms go to zero, so the total sum is 2.

Therefore, the answer is 2? But let me check again with another approach to make sure I didn't make a mistake.

Alternatively, maybe I can use generating functions or another identity. But given that the telescoping works out and the partial sums seem to approach 2, I think that's correct.

But let me check for n=1 to n=1: sum is 1/2. According to the formula, S_1 = 1 + 1 - 1/F_2 - 1/F_3 = 2 - 1/1 - 1/2 = 2 - 1 - 0.5 = 0.5. Correct.

For n=2: S_2 = 1 + 1 - 1/F_3 - 1/F_4 = 2 - 1/2 - 1/3 = 2 - 5/6 = 7/6 ≈ 1.166..., which matches the earlier calculation.

For n=3: S_3 = 2 - 1/F_4 - 1/F_5 = 2 - 1/3 - 1/5 = 2 - 8/15 = 22/15 ≈ 1.466..., which matches.

So yes, the formula is correct. Therefore, the infinite sum is 2.

But wait, let me check if there's any possible mistake in the telescoping. For example, maybe I missed some terms? Let me think again.

The original sum is from n=1 to infinity of [1/F_n - 1/F_{n+2}]. So when you write the partial sum up to N, it's sum_{n=1}^N [1/F_n - 1/F_{n+2}] = sum_{n=1}^N 1/F_n - sum_{n=1}^N 1/F_{n+2} = [sum_{n=1}^N 1/F_n] - [sum_{k=3}^{N+2} 1/F_k] where I substituted k = n + 2.

Therefore, the difference is [sum_{n=1}^N 1/F_n] - [sum_{k=3}^{N+2} 1/F_k] = [1/F_1 + 1/F_2 + sum_{k=3}^N 1/F_k] - [sum_{k=3}^N 1/F_k + 1/F_{N+1} + 1/F_{N+2}}] = 1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}.

Yes, that's correct. Therefore, the partial sum is indeed 2 - [1/F_{N+1} + 1/F_{N+2}], which tends to 2 as N approaches infinity.

Therefore, the answer is 2.

But let me check if there's any possible mistake in the initial step where I transformed the term.

Original term: F_{n+1}/(F_n F_{n+2})

We used the identity F_{n+2} = F_{n+1} + F_n, so F_{n+1}/(F_n F_{n+2}) = F_{n+1}/(F_n (F_{n+1} + F_n)) = [F_{n+1}/(F_n F_{n+1} + F_n^2)]? Wait, no, that's not helpful. Wait, but earlier I thought of expressing F_{n+1} as F_{n+2} - F_n. Let me check that:

From F_{n+2} = F_{n+1} + F_n, rearranged gives F_{n+1} = F_{n+2} - F_n. Therefore, substituting into the numerator:

F_{n+1}/(F_n F_{n+2}) = (F_{n+2} - F_n)/(F_n F_{n+2}) = F_{n+2}/(F_n F_{n+2}) - F_n/(F_n F_{n+2}) = 1/F_n - 1/F_{n+2}.

Yes, that's correct. So that step is valid.

Therefore, the entire reasoning is correct, and the sum is indeed 2.

But just to be thorough, let me check with another approach. Maybe using generating functions or another identity.

Alternatively, since Fibonacci numbers have generating functions, but I don't know if that would be helpful here. Alternatively, maybe using the closed-form expression for Fibonacci numbers, Binet's formula: $F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}$, where $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$.

But maybe that's overcomplicating. However, if I wanted to check the limit of 1/F_n as n approaches infinity, since |ψ| < 1, so ψ^n approaches 0, so F_n ~ φ^n / sqrt(5), so 1/F_n ~ sqrt(5)/φ^n, which tends to 0. Therefore, the terms 1/F_{N+1} and 1/F_{N+2} go to zero, so the sum is indeed 2.

Alternatively, maybe I can think about the sum as 2 - lim_{N→∞} [1/F_{N+1} + 1/F_{N+2}] = 2 - 0 = 2.

Therefore, I think the answer is 2.

But let me check with another partial sum. Let me compute up to n=4:

Sum up to n=4: 1/2 + 2/3 + 3/10 + 5/24 ≈ 0.5 + 0.666... + 0.3 + 0.208... ≈ 0.5 + 0.666 = 1.166 + 0.3 = 1.466 + 0.208 ≈ 1.674.

According to the formula, S_4 = 2 - 1/F_5 - 1/F_6 = 2 - 1/5 - 1/8 = 2 - 0.2 - 0.125 = 2 - 0.325 = 1.675. Which matches the approximate value.

Similarly, for n=5, S_5 = 2 - 1/F_6 - 1/F_7 = 2 - 1/8 - 1/13 ≈ 2 - 0.125 - 0.0769 ≈ 2 - 0.2019 ≈ 1.798, which is close to the approximate value of 1.8.

So as N increases, the partial sums approach 2.

Therefore, I think confidently that the sum is 2.

But just to make sure, let me check if there's any possible mistake in the telescoping. For example, maybe the indices are off? Let me check again:

Original sum: starting at n=1, so the first term is n=1: 1/F_1 - 1/F_3. Then n=2: 1/F_2 - 1/F_4. Then n=3: 1/F_3 - 1/F_5. So when you add them up, the -1/F_3 cancels with the +1/F_3 from n=3. Similarly, -1/F_4 cancels with +1/F_4 from n=4, etc. Therefore, the only terms that don't cancel are the first two positive terms (from n=1 and n=2) and the last two negative terms (from n=N and n=N+1). Wait, but in the partial sum S_N, the last terms are -1/F_{N+1} - 1/F_{N+2}.

Yes, so when N approaches infinity, those terms go to zero. Therefore, the total sum is 1/F_1 + 1/F_2 = 1 + 1 = 2.

Therefore, I think the answer is 2.

But just to check, maybe there's a different way to think about it. For example, if I consider the sum from n=1 to infinity of [1/F_n - 1/F_{n+2}], which is equal to sum_{n=1}^\infty 1/F_n - sum_{n=1}^\infty 1/F_{n+2} = sum_{n=1}^\infty 1/F_n - sum_{k=3}^\infty 1/F_k = [sum_{n=1}^\infty 1/F_n] - [sum_{n=1}^\infty 1/F_n - 1/F_1 - 1/F_2] = 1/F_1 + 1/F_2 = 1 + 1 = 2.

Yes, that's another way to see it. The second sum is just the first sum shifted by two indices, so subtracting them leaves the first two terms.

Therefore, this confirms again that the sum is 2.

Therefore, I think the answer is 2.

**Final Answer**
\boxed{2}

To evaluate the infinite sum

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},
$$

where $ F_n $ denotes the $ n $-th Fibonacci number with $ F_0 = 0 $ and $ F_1 = 1 $, we begin by analyzing the structure of the general term.

---

### Step 1: Use the Fibonacci Recurrence

Recall the Fibonacci recurrence:

$$
F_{n + 2} = F_{n + 1} + F_n.
$$

We can use this to rewrite the denominator:

$$
F_n F_{n + 2} = F_n (F_{n + 1} + F_n) = F_n F_{n + 1} + F_n^2.
$$

However, a more useful approach is to express the numerator $ F_{n + 1} $ in terms of the recurrence:

$$
F_{n + 1} = F_{n + 2} - F_n.
$$

Substituting this into the original term:

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

---

### Step 2: Rewrite the Sum

Now, the original sum becomes:

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

This is a **telescoping series**. Let's write out the first few terms to observe the cancellation:

$$
\left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \left( \frac{1}{F_4} - \frac{1}{F_6} \right) + \cdots
$$

We see that all intermediate terms cancel, leaving only the first two positive terms and the last two negative terms (which go to zero as $ n \to \infty $):

$$
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right) = \frac{1}{F_1} + \frac{1}{F_2} - \lim_{N \to \infty} \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right).
$$

---

### Step 3: Evaluate the Limit

We know:

- $ F_1 = 1 $, so $ \frac{1}{F_1} = 1 $,
- $ F_2 = 1 $, so $ \frac{1}{F_2} = 1 $,
- As $ n \to \infty $, $ F_n $ grows exponentially (since Fibonacci numbers grow like $ \phi^n $, where $ \phi = \frac{1 + \sqrt{5}}{2} $), so $ \frac{1}{F_n} \to 0 $.

Thus, the limit of the negative terms is zero:

$$
\lim_{N \to \infty} \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right) = 0.
$$

---

### Final Computation

Putting it all together:

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_1} + \frac{1}{F_2} = 1 + 1 = \boxed{2}.
$$

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to compute the infinite sum $\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}}$, 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 $n \geq 2$, with $F_0 = 0$ and $F_1 = 1$.

First, maybe I should write out the first few terms of the sequence to get a sense of what's going on. 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 the sum starts at $n = 1$, so let's compute the first few terms of the sum:

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

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

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

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

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

So adding these up: 1/2 + 2/3 + 3/10 + 5/24 + 8/65 + ...

Hmm, not sure if that helps directly. Maybe I need to find a telescoping series or some identity involving Fibonacci numbers.

I remember that Fibonacci numbers have some identities. For example, Cassini's identity: $F_{n+1}F_{n-1} - F_n^2 = (-1)^n$. Maybe that's useful? Or perhaps some other identity?

Alternatively, maybe I can express the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$ in terms of differences of reciprocals or something similar. Let me think.

First, let's recall the Fibonacci recurrence: $F_{n + 2} = F_{n + 1} + F_n$. Maybe I can use this to rewrite the denominator. Let me try:

Denominator is $F_n F_{n + 2} = F_n (F_{n + 1} + F_n) = F_n F_{n + 1} + F_n^2$. Not sure if that helps yet.

Alternatively, maybe express the term as something involving $F_{n + 1}/F_n$ or $F_{n + 1}/F_{n + 2}$. Let me think about the ratio of consecutive Fibonacci numbers.

I know that $F_{n + 1}/F_n$ approaches the golden ratio $\phi = (1 + \sqrt{5})/2$ as $n$ becomes large, but I don't know if that helps here.

Alternatively, maybe I can manipulate the term $\frac{F_{n + 1}}{F_n F_{n + 2}}$ using the recurrence relation. Let me try to express it in terms of differences.

Let me consider the identity $F_{n + 2} = F_{n + 1} + F_n$. Let me solve for $F_{n + 1}$: $F_{n + 1} = F_{n + 2} - F_n$. Wait, but substituting that into the numerator:

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

Oh! That seems promising! So the term simplifies to $\frac{1}{F_n} - \frac{1}{F_{n + 2}}$. Let me check that with an example. Let's take n=1:

Original term: $F_2/(F_1 F_3) = 1/(1*2) = 1/2$.

According to the expression $\frac{1}{F_1} - \frac{1}{F_3} = 1/1 - 1/2 = 1 - 1/2 = 1/2. Correct.

Another example, n=2: original term is 2/(1*3) = 2/3. The expression is 1/F_2 - 1/F_4 = 1/1 - 1/3 = 2/3. Correct again.

n=3: original term is 3/(2*5) = 3/10. The expression is 1/F_3 - 1/F_5 = 1/2 - 1/5 = 5/10 - 2/10 = 3/10. Correct.

Great! So the general term simplifies to $\frac{1}{F_n} - \frac{1}{F_{n + 2}}$. Therefore, the sum becomes:

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

This looks like a telescoping series! Let me write out the first few terms to see:

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

n=2: 1/F_2 - 1/F_4

n=3: 1/F_3 - 1/F_5

n=4: 1/F_4 - 1/F_6

n=5: 1/F_5 - 1/F_7

...

So when we add them up, most terms should cancel. Let's see:

Sum = (1/F_1 - 1/F_3) + (1/F_2 - 1/F_4) + (1/F_3 - 1/F_5) + (1/F_4 - 1/F_6) + ...

Indeed, the -1/F_3 cancels with +1/F_3 from the next term, similarly -1/F_4 cancels with +1/F_4, etc. So all the intermediate terms cancel, and we are left with the first few terms that don't get canceled.

Specifically, the sum telescopes to:

1/F_1 + 1/F_2 - lim_{k→∞} [1/F_{k + 1} + 1/F_{k + 2}]

Wait, let me think again. Let me write the partial sum up to N terms:

S_N = sum_{n=1}^N [1/F_n - 1/F_{n + 2}] = [1/F_1 - 1/F_3] + [1/F_2 - 1/F_4] + [1/F_3 - 1/F_5] + ... + [1/F_N - 1/F_{N + 2}]

Now, grouping the terms:

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

Wait, actually, the cancellation is between -1/F_{n + 2} and the next term's +1/F_{n + 2}? Wait, let me check:

Wait, for each term, the negative part is -1/F_{n + 2}, and the next term's positive part is +1/F_{n + 1}... Wait, no. Let me think again.

Wait, the first term is 1/F_1 - 1/F_3

Second term is 1/F_2 - 1/F_4

Third term is 1/F_3 - 1/F_5

Fourth term is 1/F_4 - 1/F_6

...

So when you add them up, the -1/F_3 cancels with the +1/F_3 in the third term. Similarly, -1/F_4 cancels with +1/F_4 in the fourth term, etc. So the partial sum S_N is:

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

Because all the intermediate terms cancel. Let me check with N=2:

S_2 = [1/F_1 - 1/F_3] + [1/F_2 - 1/F_4] = 1/F_1 + 1/F_2 - 1/F_3 - 1/F_4. Which is 1/F_1 + 1/F_2 - 1/F_{2 + 1} - 1/F_{2 + 2} = yes. Similarly for N=3: S_3 = 1/F_1 + 1/F_2 - 1/F_4 - 1/F_5. Which is 1/F_1 + 1/F_2 - 1/F_{3 + 1} - 1/F_{3 + 2}. So yes, general formula is:

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 = lim_{N→∞} [1/F_1 + 1/F_2 - 1/F_{N + 1} - 1/F_{N + 2}]

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

Now, we need to evaluate the limit of -1/F_{N + 1} - 1/F_{N + 2} as N approaches infinity. Since Fibonacci numbers grow exponentially (they grow like φ^n / sqrt(5)), so 1/F_{N + 1} and 1/F_{N + 2} both approach 0 as N approaches infinity. Therefore, the limit is 0 - 0 = 0.

Therefore, the total sum is 2 - 0 = 2. Wait, so the sum equals 2? Let me check with the partial sums I calculated earlier.

Earlier, the first term was 1/2, then adding 2/3 gives 1/2 + 2/3 = 7/6 ≈ 1.166..., then adding 3/10 gives 7/6 + 3/10 = 35/30 + 9/30 = 44/30 ≈ 1.466..., then adding 5/24 gives 44/30 + 5/24 = (176 + 25)/120 = 201/120 ≈ 1.675, then adding 8/65 gives 201/120 + 8/65 = let's compute that:

Convert to common denominator, which is 120*65 = 7800. 201/120 = (201*65)/7800 = let me compute 201*65:

200*65 = 13,000; 1*65 = 65; total 13,065. Then 8/65 = (8*120)/7800 = 960/7800. So total is 13,065 + 960 = 14,025 / 7800 ≈ 1.8.

So after 5 terms, the partial sum is about 1.8, which is approaching 2. So that seems plausible.

Therefore, the sum converges to 2.

Wait, but let me check again the telescoping sum. The partial sum S_N is 1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2} = 1 + 1 - [1/F_{N+1} + 1/F_{N+2}]. So as N approaches infinity, the subtracted terms go to zero, so the total sum is 2.

Therefore, the answer is 2? But let me check again with another approach to make sure I didn't make a mistake.

Alternatively, maybe I can use generating functions or another identity. But given that the telescoping works out and the partial sums seem to approach 2, I think that's correct.

But let me check for n=1 to n=1: sum is 1/2. According to the formula, S_1 = 1 + 1 - 1/F_2 - 1/F_3 = 2 - 1/1 - 1/2 = 2 - 1 - 0.5 = 0.5. Correct.

For n=2: S_2 = 1 + 1 - 1/F_3 - 1/F_4 = 2 - 1/2 - 1/3 = 2 - 5/6 = 7/6 ≈ 1.166..., which matches the earlier calculation.

For n=3: S_3 = 2 - 1/F_4 - 1/F_5 = 2 - 1/3 - 1/5 = 2 - 8/15 = 22/15 ≈ 1.466..., which matches.

So yes, the formula is correct. Therefore, the infinite sum is 2.

But wait, let me check if there's any possible mistake in the telescoping. For example, maybe I missed some terms? Let me think again.

The original sum is from n=1 to infinity of [1/F_n - 1/F_{n+2}]. So when you write the partial sum up to N, it's sum_{n=1}^N [1/F_n - 1/F_{n+2}] = sum_{n=1}^N 1/F_n - sum_{n=1}^N 1/F_{n+2} = [sum_{n=1}^N 1/F_n] - [sum_{k=3}^{N+2} 1/F_k] where I substituted k = n + 2.

Therefore, the difference is [sum_{n=1}^N 1/F_n] - [sum_{k=3}^{N+2} 1/F_k] = [1/F_1 + 1/F_2 + sum_{k=3}^N 1/F_k] - [sum_{k=3}^N 1/F_k + 1/F_{N+1} + 1/F_{N+2}}] = 1/F_1 + 1/F_2 - 1/F_{N+1} - 1/F_{N+2}.

Yes, that's correct. Therefore, the partial sum is indeed 2 - [1/F_{N+1} + 1/F_{N+2}], which tends to 2 as N approaches infinity.

Therefore, the answer is 2.

But let me check if there's any possible mistake in the initial step where I transformed the term.

Original term: F_{n+1}/(F_n F_{n+2})

We used the identity F_{n+2} = F_{n+1} + F_n, so F_{n+1}/(F_n F_{n+2}) = F_{n+1}/(F_n (F_{n+1} + F_n)) = [F_{n+1}/(F_n F_{n+1} + F_n^2)]? Wait, no, that's not helpful. Wait, but earlier I thought of expressing F_{n+1} as F_{n+2} - F_n. Let me check that:

From F_{n+2} = F_{n+1} + F_n, rearranged gives F_{n+1} = F_{n+2} - F_n. Therefore, substituting into the numerator:

F_{n+1}/(F_n F_{n+2}) = (F_{n+2} - F_n)/(F_n F_{n+2}) = F_{n+2}/(F_n F_{n+2}) - F_n/(F_n F_{n+2}) = 1/F_n - 1/F_{n+2}.

Yes, that's correct. So that step is valid.

Therefore, the entire reasoning is correct, and the sum is indeed 2.

But just to be thorough, let me check with another approach. Maybe using generating functions or another identity.

Alternatively, since Fibonacci numbers have generating functions, but I don't know if that would be helpful here. Alternatively, maybe using the closed-form expression for Fibonacci numbers, Binet's formula: $F_n = \frac{\phi^n - \psi^n}{\sqrt{5}}$, where $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$.

But maybe that's overcomplicating. However, if I wanted to check the limit of 1/F_n as n approaches infinity, since |ψ| < 1, so ψ^n approaches 0, so F_n ~ φ^n / sqrt(5), so 1/F_n ~ sqrt(5)/φ^n, which tends to 0. Therefore, the terms 1/F_{N+1} and 1/F_{N+2} go to zero, so the sum is indeed 2.

Alternatively, maybe I can think about the sum as 2 - lim_{N→∞} [1/F_{N+1} + 1/F_{N+2}] = 2 - 0 = 2.

Therefore, I think the answer is 2.

But let me check with another partial sum. Let me compute up to n=4:

Sum up to n=4: 1/2 + 2/3 + 3/10 + 5/24 ≈ 0.5 + 0.666... + 0.3 + 0.208... ≈ 0.5 + 0.666 = 1.166 + 0.3 = 1.466 + 0.208 ≈ 1.674.

According to the formula, S_4 = 2 - 1/F_5 - 1/F_6 = 2 - 1/5 - 1/8 = 2 - 0.2 - 0.125 = 2 - 0.325 = 1.675. Which matches the approximate value.

Similarly, for n=5, S_5 = 2 - 1/F_6 - 1/F_7 = 2 - 1/8 - 1/13 ≈ 2 - 0.125 - 0.0769 ≈ 2 - 0.2019 ≈ 1.798, which is close to the approximate value of 1.8.

So as N increases, the partial sums approach 2.

Therefore, I think confidently that the sum is 2.

But just to make sure, let me check if there's any possible mistake in the telescoping. For example, maybe the indices are off? Let me check again:

Original sum: starting at n=1, so the first term is n=1: 1/F_1 - 1/F_3. Then n=2: 1/F_2 - 1/F_4. Then n=3: 1/F_3 - 1/F_5. So when you add them up, the -1/F_3 cancels with the +1/F_3 from n=3. Similarly, -1/F_4 cancels with +1/F_4 from n=4, etc. Therefore, the only terms that don't cancel are the first two positive terms (from n=1 and n=2) and the last two negative terms (from n=N and n=N+1). Wait, but in the partial sum S_N, the last terms are -1/F_{N+1} - 1/F_{N+2}.

Yes, so when N approaches infinity, those terms go to zero. Therefore, the total sum is 1/F_1 + 1/F_2 = 1 + 1 = 2.

Therefore, I think the answer is 2.

But just to check, maybe there's a different way to think about it. For example, if I consider the sum from n=1 to infinity of [1/F_n - 1/F_{n+2}], which is equal to sum_{n=1}^\infty 1/F_n - sum_{n=1}^\infty 1/F_{n+2} = sum_{n=1}^\infty 1/F_n - sum_{k=3}^\infty 1/F_k = [sum_{n=1}^\infty 1/F_n] - [sum_{n=1}^\infty 1/F_n - 1/F_1 - 1/F_2] = 1/F_1 + 1/F_2 = 1 + 1 = 2.

Yes, that's another way to see it. The second sum is just the first sum shifted by two indices, so subtracting them leaves the first two terms.

Therefore, this confirms again that the sum is 2.

Therefore, I think the answer is 2.

**Final Answer**
\boxed{2}

To evaluate the infinite sum

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}},
$$

where $ F_n $ denotes the $ n $-th Fibonacci number with $ F_0 = 0 $ and $ F_1 = 1 $, we begin by analyzing the structure of the general term.

---

### Step 1: Use the Fibonacci Recurrence

Recall the Fibonacci recurrence:

$$
F_{n + 2} = F_{n + 1} + F_n.
$$

We can use this to rewrite the denominator:

$$
F_n F_{n + 2} = F_n (F_{n + 1} + F_n) = F_n F_{n + 1} + F_n^2.
$$

However, a more useful approach is to express the numerator $ F_{n + 1} $ in terms of the recurrence:

$$
F_{n + 1} = F_{n + 2} - F_n.
$$

Substituting this into the original term:

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

---

### Step 2: Rewrite the Sum

Now, the original sum becomes:

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

This is a **telescoping series**. Let's write out the first few terms to observe the cancellation:

$$
\left( \frac{1}{F_1} - \frac{1}{F_3} \right) + \left( \frac{1}{F_2} - \frac{1}{F_4} \right) + \left( \frac{1}{F_3} - \frac{1}{F_5} \right) + \left( \frac{1}{F_4} - \frac{1}{F_6} \right) + \cdots
$$

We see that all intermediate terms cancel, leaving only the first two positive terms and the last two negative terms (which go to zero as $ n \to \infty $):

$$
\sum_{n = 1}^\infty \left( \frac{1}{F_n} - \frac{1}{F_{n + 2}} \right) = \frac{1}{F_1} + \frac{1}{F_2} - \lim_{N \to \infty} \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right).
$$

---

### Step 3: Evaluate the Limit

We know:

- $ F_1 = 1 $, so $ \frac{1}{F_1} = 1 $,
- $ F_2 = 1 $, so $ \frac{1}{F_2} = 1 $,
- As $ n \to \infty $, $ F_n $ grows exponentially (since Fibonacci numbers grow like $ \phi^n $, where $ \phi = \frac{1 + \sqrt{5}}{2} $), so $ \frac{1}{F_n} \to 0 $.

Thus, the limit of the negative terms is zero:

$$
\lim_{N \to \infty} \left( \frac{1}{F_{N + 1}} + \frac{1}{F_{N + 2}} \right) = 0.
$$

---

### Final Computation

Putting it all together:

$$
\sum_{n = 1}^\infty \frac{F_{n + 1}}{F_n F_{n + 2}} = \frac{1}{F_1} + \frac{1}{F_2} = 1 + 1 = \boxed{2}.
$$

---

### 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*}


Explanation:
Okay, so I need to find the value of $ x^3 - 2x + 1 $ given that $ x^2 - x - 1 = 0 $. Hmm, let me think about how to approach this.

First, I remember that if I have an equation involving $ x $, I can sometimes use substitution or manipulate the equation to express higher powers of $ x $ in terms of lower ones. Since the given equation is quadratic, maybe I can solve for $ x^2 $ and then use that to find $ x^3 $?

Let me start by solving the quadratic equation $ x^2 - x - 1 = 0 $. Using the quadratic formula, $ x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} $. But maybe I don't need to find the exact value of $ x $; perhaps I can work algebraically with the equation.

The original equation is $ x^2 = x + 1 $. That seems useful. If I can express $ x^3 $ in terms of lower powers of $ x $, then I can substitute into the expression $ x^3 - 2x + 1 $.

Let me try multiplying both sides of the equation $ x^2 = x + 1 $ by $ x $ to get $ x^3 $.

Multiplying both sides by $ x $:

$ x^3 = x(x + 1) = x^2 + x $.

But wait, I know from the original equation that $ x^2 = x + 1 $, so I can substitute that into the expression for $ x^3 $:

$ x^3 = (x + 1) + x = 2x + 1 $.

Okay, so $ x^3 = 2x + 1 $. Now, substitute this into the original expression $ x^3 - 2x + 1 $:

$ (2x + 1) - 2x + 1 $.

Let me simplify that:

First, distribute the terms: $ 2x + 1 - 2x + 1 $.

Combine like terms: $ (2x - 2x) + (1 + 1) = 0 + 2 = 2 $.

Wait, so does that mean the value is 2? Let me check my steps again to make sure I didn't make a mistake.

Starting with $ x^2 = x + 1 $. Multiply both sides by $ x $: $ x^3 = x^2 + x $. Then substitute $ x^2 $ with $ x + 1 $: $ x^3 = (x + 1) + x = 2x + 1 $. Then plug into $ x^3 - 2x + 1 $: $ 2x + 1 - 2x + 1 $. The $ 2x $ and $ -2x $ cancel out, leaving $ 1 + 1 = 2 $. That seems correct.

But let me verify by plugging in the actual roots. Let me compute $ x^3 - 2x + 1 $ for $ x = \frac{1 + \sqrt{5}}{2} $ and see if it equals 2.

First, compute $ x = \frac{1 + \sqrt{5}}{2} $. Let me calculate $ x^3 $:

Alternatively, maybe I can use the original equation to compute $ x^3 $. Wait, I already did that algebraically, but let me check numerically.

Let me compute $ x \approx \frac{1 + 2.236}{2} \approx \frac{3.236}{2} \approx 1.618 $.

Compute $ x^3 $: $ (1.618)^3 $. Let me calculate that:

First, $ 1.618^2 \approx 2.618 $, then multiply by 1.618: 2.618 * 1.618 ≈ 4.236.

Then compute $ -2x + 1 $: -2*(1.618) + 1 ≈ -3.236 + 1 ≈ -2.236.

So total expression: 4.236 - 2.236 ≈ 2.0. That checks out! So the value is indeed 2.

Alternatively, let me check with the other root $ x = \frac{1 - \sqrt{5}}{2} \approx \frac{1 - 2.236}{2} \approx \frac{-1.236}{2} \approx -0.618 $.

Compute $ x^3 - 2x + 1 $:

First, $ x^3 \approx (-0.618)^3 ≈ -0.236 $.

Then, -2x + 1 ≈ -2*(-0.618) + 1 ≈ 1.236 + 1 ≈ 2.236.

Adding them together: -0.236 + 2.236 ≈ 2.0. So that also gives 2.

Therefore, regardless of which root we take, the value is 2.

So my initial algebraic approach was correct. Therefore, the answer is 2.

But just to make sure I didn't skip any steps or make any miscalculations, let me go through the algebra again:

Given $ x^2 = x + 1 $.

Multiply both sides by x: $ x^3 = x^2 + x $.

Replace $ x^2 $ with $ x + 1 $: $ x^3 = (x + 1) + x = 2x + 1 $.

Therefore, $ x^3 - 2x + 1 = (2x + 1) - 2x + 1 = 2x + 1 - 2x + 1 = 2 $.

Yes, that's straightforward. So the answer is 2.

Alternatively, another way to think about it is to express the original expression in terms of the given equation. Since we know $ x^2 = x + 1 $, maybe we can write $ x^3 - 2x + 1 $ as $ x \cdot x^2 - 2x + 1 $, and substitute $ x^2 $:

So $ x \cdot (x + 1) - 2x + 1 = x^2 + x - 2x + 1 = x^2 - x + 1 $.

But wait, from the original equation, $ x^2 - x - 1 = 0 $, so $ x^2 - x = 1 $. Therefore, substituting into the above expression:

$ x^2 - x + 1 = 1 + 1 = 2 $.

Oh, that's another way to get the same result. So that's a more straightforward substitution.

So, breaking it down:

Original expression: $ x^3 - 2x + 1 $.

Express $ x^3 $ as $ x \cdot x^2 $, which is $ x(x + 1) $, leading to $ x^2 + x $, then substitute $ x^2 $ again. But another way is to write $ x^3 - 2x + 1 = x \cdot x^2 - 2x + 1 $, then substitute $ x^2 = x + 1 $, getting $ x(x + 1) - 2x + 1 $, which is $ x^2 + x - 2x + 1 = x^2 - x + 1 $. Then since $ x^2 - x = 1 $, we have $ 1 + 1 = 2 $.

Either way, the answer is 2.

Therefore, I can be confident that the value of $ x^3 - 2x + 1 $ is 2.

**Final Answer**
\boxed{2}

To evaluate the expression $ x^3 - 2x + 1 $, given that $ x^2 - x - 1 = 0 $, we can use algebraic manipulation based on the given equation.

---

### Step 1: Use the given equation to express $ x^2 $

From the equation:

$$
x^2 - x - 1 = 0 \quad \Rightarrow \quad x^2 = x + 1
$$

This allows us to express higher powers of $ x $ in terms of lower powers.

---

### Step 2: Express $ x^3 $ in terms of $ x $

Multiply both sides of the equation $ x^2 = x + 1 $ by $ x $:

$$
x^3 = x(x + 1) = x^2 + x
$$

Now substitute $ x^2 = x + 1 $ into the expression:

$$
x^3 = (x + 1) + x = 2x + 1
$$

---

### Step 3: Substitute into the target expression

We are asked to find the value of:

$$
x^3 - 2x + 1
$$

Substitute $ x^3 = 2x + 1 $:

$$
x^3 - 2x + 1 = (2x + 1) - 2x + 1
$$

Simplify:

$$
(2x - 2x) + (1 + 1) = 0 + 2 = 2
$$

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ x^3 - 2x + 1 $ given that $ x^2 - x - 1 = 0 $. Hmm, let me think about how to approach this.

First, I remember that if I have an equation involving $ x $, I can sometimes use substitution or manipulate the equation to express higher powers of $ x $ in terms of lower ones. Since the given equation is quadratic, maybe I can solve for $ x^2 $ and then use that to find $ x^3 $?

Let me start by solving the quadratic equation $ x^2 - x - 1 = 0 $. Using the quadratic formula, $ x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} $. But maybe I don't need to find the exact value of $ x $; perhaps I can work algebraically with the equation.

The original equation is $ x^2 = x + 1 $. That seems useful. If I can express $ x^3 $ in terms of lower powers of $ x $, then I can substitute into the expression $ x^3 - 2x + 1 $.

Let me try multiplying both sides of the equation $ x^2 = x + 1 $ by $ x $ to get $ x^3 $.

Multiplying both sides by $ x $:

$ x^3 = x(x + 1) = x^2 + x $.

But wait, I know from the original equation that $ x^2 = x + 1 $, so I can substitute that into the expression for $ x^3 $:

$ x^3 = (x + 1) + x = 2x + 1 $.

Okay, so $ x^3 = 2x + 1 $. Now, substitute this into the original expression $ x^3 - 2x + 1 $:

$ (2x + 1) - 2x + 1 $.

Let me simplify that:

First, distribute the terms: $ 2x + 1 - 2x + 1 $.

Combine like terms: $ (2x - 2x) + (1 + 1) = 0 + 2 = 2 $.

Wait, so does that mean the value is 2? Let me check my steps again to make sure I didn't make a mistake.

Starting with $ x^2 = x + 1 $. Multiply both sides by $ x $: $ x^3 = x^2 + x $. Then substitute $ x^2 $ with $ x + 1 $: $ x^3 = (x + 1) + x = 2x + 1 $. Then plug into $ x^3 - 2x + 1 $: $ 2x + 1 - 2x + 1 $. The $ 2x $ and $ -2x $ cancel out, leaving $ 1 + 1 = 2 $. That seems correct.

But let me verify by plugging in the actual roots. Let me compute $ x^3 - 2x + 1 $ for $ x = \frac{1 + \sqrt{5}}{2} $ and see if it equals 2.

First, compute $ x = \frac{1 + \sqrt{5}}{2} $. Let me calculate $ x^3 $:

Alternatively, maybe I can use the original equation to compute $ x^3 $. Wait, I already did that algebraically, but let me check numerically.

Let me compute $ x \approx \frac{1 + 2.236}{2} \approx \frac{3.236}{2} \approx 1.618 $.

Compute $ x^3 $: $ (1.618)^3 $. Let me calculate that:

First, $ 1.618^2 \approx 2.618 $, then multiply by 1.618: 2.618 * 1.618 ≈ 4.236.

Then compute $ -2x + 1 $: -2*(1.618) + 1 ≈ -3.236 + 1 ≈ -2.236.

So total expression: 4.236 - 2.236 ≈ 2.0. That checks out! So the value is indeed 2.

Alternatively, let me check with the other root $ x = \frac{1 - \sqrt{5}}{2} \approx \frac{1 - 2.236}{2} \approx \frac{-1.236}{2} \approx -0.618 $.

Compute $ x^3 - 2x + 1 $:

First, $ x^3 \approx (-0.618)^3 ≈ -0.236 $.

Then, -2x + 1 ≈ -2*(-0.618) + 1 ≈ 1.236 + 1 ≈ 2.236.

Adding them together: -0.236 + 2.236 ≈ 2.0. So that also gives 2.

Therefore, regardless of which root we take, the value is 2.

So my initial algebraic approach was correct. Therefore, the answer is 2.

But just to make sure I didn't skip any steps or make any miscalculations, let me go through the algebra again:

Given $ x^2 = x + 1 $.

Multiply both sides by x: $ x^3 = x^2 + x $.

Replace $ x^2 $ with $ x + 1 $: $ x^3 = (x + 1) + x = 2x + 1 $.

Therefore, $ x^3 - 2x + 1 = (2x + 1) - 2x + 1 = 2x + 1 - 2x + 1 = 2 $.

Yes, that's straightforward. So the answer is 2.

Alternatively, another way to think about it is to express the original expression in terms of the given equation. Since we know $ x^2 = x + 1 $, maybe we can write $ x^3 - 2x + 1 $ as $ x \cdot x^2 - 2x + 1 $, and substitute $ x^2 $:

So $ x \cdot (x + 1) - 2x + 1 = x^2 + x - 2x + 1 = x^2 - x + 1 $.

But wait, from the original equation, $ x^2 - x - 1 = 0 $, so $ x^2 - x = 1 $. Therefore, substituting into the above expression:

$ x^2 - x + 1 = 1 + 1 = 2 $.

Oh, that's another way to get the same result. So that's a more straightforward substitution.

So, breaking it down:

Original expression: $ x^3 - 2x + 1 $.

Express $ x^3 $ as $ x \cdot x^2 $, which is $ x(x + 1) $, leading to $ x^2 + x $, then substitute $ x^2 $ again. But another way is to write $ x^3 - 2x + 1 = x \cdot x^2 - 2x + 1 $, then substitute $ x^2 = x + 1 $, getting $ x(x + 1) - 2x + 1 $, which is $ x^2 + x - 2x + 1 = x^2 - x + 1 $. Then since $ x^2 - x = 1 $, we have $ 1 + 1 = 2 $.

Either way, the answer is 2.

Therefore, I can be confident that the value of $ x^3 - 2x + 1 $ is 2.

**Final Answer**
\boxed{2}

To evaluate the expression $ x^3 - 2x + 1 $, given that $ x^2 - x - 1 = 0 $, we can use algebraic manipulation based on the given equation.

---

### Step 1: Use the given equation to express $ x^2 $

From the equation:

$$
x^2 - x - 1 = 0 \quad \Rightarrow \quad x^2 = x + 1
$$

This allows us to express higher powers of $ x $ in terms of lower powers.

---

### Step 2: Express $ x^3 $ in terms of $ x $

Multiply both sides of the equation $ x^2 = x + 1 $ by $ x $:

$$
x^3 = x(x + 1) = x^2 + x
$$

Now substitute $ x^2 = x + 1 $ into the expression:

$$
x^3 = (x + 1) + x = 2x + 1
$$

---

### Step 3: Substitute into the target expression

We are asked to find the value of:

$$
x^3 - 2x + 1
$$

Substitute $ x^3 = 2x + 1 $:

$$
x^3 - 2x + 1 = (2x + 1) - 2x + 1
$$

Simplify:

$$
(2x - 2x) + (1 + 1) = 0 + 2 = 2
$$

---

### Final Answer

$$
\boxed{2}
$$

Solution:
Rearranging $x^2 - x - 1= 0$ gives $x^2 = x + 1$.  So, repeatedly substituting $x+1$ for $x^2$ gives us \begin{align*}
x^3 - 2x + 1 &= x(x^2)-2x + 1\\
&=x(x+1) - 2x + 1\\
&= x^2 + x -2x + 1\\
&= x^2 - x + 1\\
&= (x+1) - x + 1\\
&=\boxed{2}
\end{align*}


Explanation:
Okay, so I need to figure out the value of 56 Drinkets in Trinkets. Let me start by understanding the relationships between the different units here.

First, the problem says that one Trinket is equal to 4 Blinkets. So, Trinket to Blinket is straightforward: 1 Trinket = 4 Blinkets. That means if I have a certain number of Trinkets, I can convert them to Blinkets by multiplying by 4. Conversely, if I have Blinkets and want to convert to Trinkets, I would divide by 4.

Then there's another conversion: 3 Blinkets are equal to 7 Drinkets. So, Blinket to Drinket is a bit more complex. Let me write down these relationships:

1 Trinket = 4 Blinkets
3 Blinkets = 7 Drinkets

So, the goal is to convert 56 Drinkets into Trinkets. To do this, I probably need to convert Drinkets to Blinkets first, and then Blinkets to Trinkets. Let me check if that makes sense.

Starting with Drinkets. If 3 Blinkets = 7 Drinkets, then to find out how many Blinkets are in 1 Drinket, I can divide both sides by 7. That would give me (3/7) Blinkets = 1 Drinket. Therefore, 1 Drinket is equal to 3/7 Blinkets.

But wait, I need to convert 56 Drinkets to Blinkets. So, if 1 Drinket is 3/7 Blinkets, then 56 Drinkets would be 56 multiplied by 3/7 Blinkets. Let me calculate that.

First, 56 divided by 7 is 8. Then, 8 multiplied by 3 is 24. So, 56 Drinkets equal 24 Blinkets. Let me verify that again. If 3 Blinkets = 7 Drinkets, then 24 Blinkets would be (24 / 3) * 7 = 8 * 7 = 56 Drinkets. Yes, that checks out. So, 56 Drinkets is equivalent to 24 Blinkets.

Now, I need to convert 24 Blinkets into Trinkets. Since 1 Trinket is 4 Blinkets, then 1 Blinket is 1/4 Trinket. Therefore, 24 Blinkets would be 24 multiplied by (1/4) Trinkets.

Calculating that: 24 divided by 4 is 6. So, 24 Blinkets equal 6 Trinkets.

Therefore, 56 Drinkets is equal to 6 Trinkets. Let me just make sure I didn't make any mistakes in the conversions.

Starting again: 56 Drinkets. Since 3 Blinkets = 7 Drinkets, then the number of Blinkets in 56 Drinkets is (56 Drinkets) * (3 Blinkets / 7 Drinkets) = (56 * 3)/7 Blinkets. 56 divided by 7 is 8, 8*3 is 24. So that's 24 Blinkets. Then converting Blinkets to Trinkets: 24 Blinkets * (1 Trinket / 4 Blinkets) = 24/4 = 6 Trinkets. Yep, that seems consistent.

Alternatively, maybe I can set up a conversion chain from Drinkets to Trinkets directly. Let me try that.

We know that 3 Blinkets = 7 Drinkets, so we can write Blinkets in terms of Drinkets: 1 Blinket = 7/3 Drinkets. Wait, no. Wait, if 3 Blinkets = 7 Drinkets, then 1 Blinket = 7/3 Drinkets? Wait, that would mean that Blinkets are larger than Drinkets? Wait, no. Wait, if 3 Blinkets equal 7 Drinkets, then 1 Blinket equals 7/3 Drinkets. So, Blinkets are worth more than Drinkets? Wait, that seems counterintuitive. Wait, if you have 3 Blinkets, they are equal to 7 Drinkets. So, each Blinket is worth more than a Drinket. So, 1 Blinket = 7/3 Drinkets ≈ 2.333 Drinkets.

But if I want to convert Drinkets to Blinkets, then 1 Drinket is 3/7 Blinkets, which is approximately 0.428 Blinkets. Which matches with the previous calculation.

So, if I want to convert Drinkets to Trinkets, I can first convert Drinkets to Blinkets, then Blinkets to Trinkets.

Alternatively, maybe I can find a direct conversion from Drinkets to Trinkets by combining the two ratios. Let me try that.

We have:

1 Trinket = 4 Blinkets
3 Blinkets = 7 Drinkets

So, let me express Trinkets in terms of Drinkets.

First, from the second equation: 3 Blinkets = 7 Drinkets => 1 Blinket = 7/3 Drinkets.

Then, substitute into the first equation: 1 Trinket = 4 Blinkets = 4*(7/3 Drinkets) = 28/3 Drinkets.

Therefore, 1 Trinket = 28/3 Drinkets.

Therefore, to find how many Trinkets are in 56 Drinkets, we can divide 56 by (28/3).

Which is 56 * (3/28) = (56/28)*3 = 2*3 = 6.

So, again, 56 Drinkets equals 6 Trinkets.

So, both methods give me the same answer, which is reassuring.

Alternatively, maybe I can use dimensional analysis. Let me write that out.

Starting with 56 Drinkets, and I want to convert to Trinkets.

So, 56 Drinkets * (3 Blinkets / 7 Drinkets) * (1 Trinket / 4 Blinkets).

Let me compute that.

First, the Drinkets unit cancels out: 56 Drinkets * (3 Blinkets / 7 Drinkets) = 56 * 3 / 7 Blinkets.

Then, multiplying by (1 Trinket / 4 Blinkets): (56 * 3 / 7) * (1 / 4) Trinkets.

Calculating the numbers: 56 divided by 7 is 8, 8 times 3 is 24, 24 divided by 4 is 6.

So, again, 6 Trinkets.

So, all methods lead to the same answer. Therefore, I can be confident that 56 Drinkets is equal to 6 Trinkets.

But just to make sure, let me think if there's another way someone might approach this problem and maybe make a mistake. For example, if someone tried to convert Trinkets to Drinkets first, but since we need the reverse, that might complicate things.

Alternatively, someone might miscalculate the conversion factors. For instance, if they thought that 3 Blinkets = 7 Drinkets implies 1 Drinket = 3/7 Blinkets, which is correct, but if they messed up the multiplication or division steps, they might get a wrong answer.

But in my calculations, I checked each step, and they all seem consistent.

Another check: Let's see how many Drinkets are in 1 Trinket.

From above, 1 Trinket = 4 Blinkets. And 3 Blinkets = 7 Drinkets. So, 4 Blinkets = (4/3)*7 Drinkets = 28/3 Drinkets ≈ 9.333 Drinkets.

Therefore, 1 Trinket ≈ 9.333 Drinkets.

Therefore, 56 Drinkets would be 56 divided by (28/3) = 56 * 3 /28 = 6 Trinkets.

Yes, same result.

Alternatively, if I think in terms of ratios.

Let me think of the conversion rates as fractions.

We have:

Trinket to Blinket: 1 T = 4 B => T/B = 4 => B/T = 1/4

Blinket to Drinket: 3 B = 7 D => B/D = 7/3 => D/B = 3/7

Therefore, to get from D to T, we can do D to B to T.

So, D to B is D * (B/D) = D * (7/3) ? Wait, no. Wait, if 3 B = 7 D, then D = (3/7) B. Therefore, to convert D to B, you multiply by (3/7). Wait, no. Wait, if D = (3/7) B, then B = (7/3) D.

Wait, let me clarify.

If 3 B = 7 D, then solving for B gives B = (7/3) D. Therefore, to convert D to B, you multiply by 7/3.

Similarly, to convert B to T, since 1 T = 4 B, then T = B / 4.

Therefore, combining these two steps:

Starting with D, multiply by 7/3 to get B, then divide by 4 to get T.

So, overall, D * (7/3) * (1/4) = D * 7/(12).

Wait, but that would mean that 1 D = 7/12 T.

Wait, but earlier we had 1 T = 28/3 D, so 1 D = 3/28 T.

Wait, there's a discrepancy here. Which is correct?

Wait, let me check.

If 1 T = 4 B, and 3 B = 7 D, then 1 B = 7/3 D. Therefore, 1 T = 4*(7/3 D) = 28/3 D. Therefore, 1 D = 3/28 T.

But according to the other approach, if I convert D to B by multiplying by 7/3, then B to T by multiplying by 1/4, then total factor is 7/3 * 1/4 = 7/12. Therefore, 1 D = 7/12 T?

But that contradicts the previous result.

Wait, so which one is correct?

Wait, there must be a mistake here. Let me check again.

If 3 B = 7 D, then 1 D = 3/7 B.

Therefore, to convert D to B, you multiply by 3/7.

But earlier I thought that B = 7/3 D. Which is correct?

Let me take numbers. Suppose I have 3 B = 7 D.

So, if I have 7 D, that's equal to 3 B. Therefore, 1 D is equal to 3/7 B.

Therefore, to convert D to B, you multiply by 3/7.

But if I have 1 B, that's equal to 7/3 D.

Therefore, if I have 1 D, it's 3/7 B.

Therefore, if I want to convert D to T, I need to first convert D to B, which is multiplying by 3/7, then convert B to T, which is multiplying by 1/4.

Therefore, total factor is 3/7 * 1/4 = 3/28.

Therefore, 1 D = 3/28 T.

Therefore, 56 D = 56 * 3/28 T.

Which is 56 divided by 28 is 2, 2 * 3 = 6.

So, 6 T.

But earlier, when I thought of converting D to B by multiplying by 7/3, that was wrong.

So, where did that mistake come from?

Ah, I think I confused the direction of conversion.

If 3 B = 7 D, then to get B in terms of D, you have B = (7/3) D.

But that would mean that if you have D, multiplying by 7/3 gives you B.

But that would be if you have D and want to get B.

Wait, let me think.

If 3 B = 7 D, then solving for B gives B = (7/3) D.

Therefore, if you have D, multiplying by 7/3 gives you B.

But that would mean that 1 D = 7/3 B?

Wait, no. Wait, if B = (7/3) D, then 1 D = 3/7 B.

Wait, this is confusing. Let me use algebra.

Given 3 B = 7 D.

Solve for B: B = (7/3) D.

Therefore, if you have D, multiplying by 7/3 gives you B.

But that would mean that 1 D = 7/3 B? No, that can't be.

Wait, if B = (7/3) D, then D = (3/7) B.

So, if you have D, you can get B by multiplying D by (7/3).

Wait, let me take numbers. Suppose D = 7. Then, according to the equation, 3 B = 7 D = 7*7 = 49. Therefore, B = 49 / 3 ≈ 16.333.

But according to B = (7/3) D, if D = 7, then B = (7/3)*7 ≈ 16.333. Which matches.

Therefore, if you have D, multiplying by 7/3 gives you B.

But if you have 1 D, then B = 7/3 D.

But that would mean that 1 D is equal to 7/3 B? No, that would be if you had B = 7/3 D, then D = 3/7 B.

So, if you have 1 D, it's equal to 3/7 B.

But if you have 1 B, it's equal to 7/3 D.

Therefore, converting D to B requires multiplying by 3/7, and converting B to D requires multiplying by 7/3.

Therefore, earlier when I thought of converting D to B by multiplying by 7/3, that was wrong. It should be multiplying by 3/7.

So, that was my mistake earlier.

So, going back, when I did the dimensional analysis earlier, I had:

56 Drinkets * (3 Blinkets / 7 Drinkets) * (1 Trinket / 4 Blinkets)

Which is correct because 3 Blinkets / 7 Drinkets is the conversion factor from Drinkets to Blinkets (since 3 Blinkets = 7 Drinkets, so 1 Drinket = 3/7 Blinkets).

Therefore, multiplying by 3/7 gives Blinkets. Then multiplying by 1/4 gives Trinkets.

So, that was correct.

But when I tried to think of converting D to B by multiplying by 7/3, that was wrong. It should be multiplying by 3/7.

So, that was a confusion in the direction of the conversion.

Therefore, the correct conversion factor from D to B is 3/7, not 7/3.

Therefore, the earlier calculation where I had 56 Drinkets * 3/7 Blinkets per Drinket * 1/4 Trinket per Blinket is correct.

So, that gives 56 * 3/7 * 1/4 = 6 Trinkets.

Therefore, the answer is 6 Trinkets.

So, after correcting my earlier confusion, I still arrive at the same answer.

Another way to think about it is to find the exchange rate between Drinkets and Trinkets directly.

We know that 1 Trinket = 4 Blinkets, and 3 Blinkets = 7 Drinkets.

So, let's find how many Drinkets are in 1 Trinket.

Since 1 Trinket = 4 Blinkets, and 3 Blinkets = 7 Drinkets, then 4 Blinkets = (4/3)*7 Drinkets = 28/3 Drinkets.

Therefore, 1 Trinket = 28/3 Drinkets.

Therefore, 1 Drinket = 3/28 Trinkets.

Therefore, 56 Drinkets = 56 * (3/28) Trinkets.

Calculating that: 56 divided by 28 is 2, 2 times 3 is 6.

So, again, 6 Trinkets.

Therefore, regardless of the method, the answer is 6.

I think this is solid. I considered multiple approaches, checked for consistency, corrected my earlier mistake in conversion factors, and all paths lead to the same answer.

Therefore, I can confidently say that 56 Drinkets is equal to 6 Trinkets.

**Final Answer**
The value of 56 Drinkets in Trinkets is \boxed{6}.

To determine the value of **56 Drinkets** in **Trinkets**, we need to use the given conversion relationships:

- **1 Trinket = 4 Blinkets**
- **3 Blinkets = 7 Drinkets**

---

### Step 1: Convert Drinkets to Blinkets

We are given that **3 Blinkets = 7 Drinkets**, so:

$$
1 \text{ Drinket} = \frac{3}{7} \text{ Blinkets}
$$

Now, to find how many Blinkets are in **56 Drinkets**:

$$
56 \text{ Drinkets} \times \frac{3}{7} \text{ Blinkets per Drinket} = \frac{56 \times 3}{7} = \frac{168}{7} = 24 \text{ Blinkets}
$$

---

### Step 2: Convert Blinkets to Trinkets

We are also given that **1 Trinket = 4 Blinkets**, so:

$$
1 \text{ Blinket} = \frac{1}{4} \text{ Trinket}
$$

Now, convert **24 Blinkets** to Trinkets:

$$
24 \text{ Blinkets} \times \frac{1}{4} \text{ Trinket per Blinket} = \frac{24}{4} = 6 \text{ Trinkets}
$$

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of 56 Drinkets in Trinkets. Let me start by understanding the relationships between the different units here.

First, the problem says that one Trinket is equal to 4 Blinkets. So, Trinket to Blinket is straightforward: 1 Trinket = 4 Blinkets. That means if I have a certain number of Trinkets, I can convert them to Blinkets by multiplying by 4. Conversely, if I have Blinkets and want to convert to Trinkets, I would divide by 4.

Then there's another conversion: 3 Blinkets are equal to 7 Drinkets. So, Blinket to Drinket is a bit more complex. Let me write down these relationships:

1 Trinket = 4 Blinkets
3 Blinkets = 7 Drinkets

So, the goal is to convert 56 Drinkets into Trinkets. To do this, I probably need to convert Drinkets to Blinkets first, and then Blinkets to Trinkets. Let me check if that makes sense.

Starting with Drinkets. If 3 Blinkets = 7 Drinkets, then to find out how many Blinkets are in 1 Drinket, I can divide both sides by 7. That would give me (3/7) Blinkets = 1 Drinket. Therefore, 1 Drinket is equal to 3/7 Blinkets.

But wait, I need to convert 56 Drinkets to Blinkets. So, if 1 Drinket is 3/7 Blinkets, then 56 Drinkets would be 56 multiplied by 3/7 Blinkets. Let me calculate that.

First, 56 divided by 7 is 8. Then, 8 multiplied by 3 is 24. So, 56 Drinkets equal 24 Blinkets. Let me verify that again. If 3 Blinkets = 7 Drinkets, then 24 Blinkets would be (24 / 3) * 7 = 8 * 7 = 56 Drinkets. Yes, that checks out. So, 56 Drinkets is equivalent to 24 Blinkets.

Now, I need to convert 24 Blinkets into Trinkets. Since 1 Trinket is 4 Blinkets, then 1 Blinket is 1/4 Trinket. Therefore, 24 Blinkets would be 24 multiplied by (1/4) Trinkets.

Calculating that: 24 divided by 4 is 6. So, 24 Blinkets equal 6 Trinkets.

Therefore, 56 Drinkets is equal to 6 Trinkets. Let me just make sure I didn't make any mistakes in the conversions.

Starting again: 56 Drinkets. Since 3 Blinkets = 7 Drinkets, then the number of Blinkets in 56 Drinkets is (56 Drinkets) * (3 Blinkets / 7 Drinkets) = (56 * 3)/7 Blinkets. 56 divided by 7 is 8, 8*3 is 24. So that's 24 Blinkets. Then converting Blinkets to Trinkets: 24 Blinkets * (1 Trinket / 4 Blinkets) = 24/4 = 6 Trinkets. Yep, that seems consistent.

Alternatively, maybe I can set up a conversion chain from Drinkets to Trinkets directly. Let me try that.

We know that 3 Blinkets = 7 Drinkets, so we can write Blinkets in terms of Drinkets: 1 Blinket = 7/3 Drinkets. Wait, no. Wait, if 3 Blinkets = 7 Drinkets, then 1 Blinket = 7/3 Drinkets? Wait, that would mean that Blinkets are larger than Drinkets? Wait, no. Wait, if 3 Blinkets equal 7 Drinkets, then 1 Blinket equals 7/3 Drinkets. So, Blinkets are worth more than Drinkets? Wait, that seems counterintuitive. Wait, if you have 3 Blinkets, they are equal to 7 Drinkets. So, each Blinket is worth more than a Drinket. So, 1 Blinket = 7/3 Drinkets ≈ 2.333 Drinkets.

But if I want to convert Drinkets to Blinkets, then 1 Drinket is 3/7 Blinkets, which is approximately 0.428 Blinkets. Which matches with the previous calculation.

So, if I want to convert Drinkets to Trinkets, I can first convert Drinkets to Blinkets, then Blinkets to Trinkets.

Alternatively, maybe I can find a direct conversion from Drinkets to Trinkets by combining the two ratios. Let me try that.

We have:

1 Trinket = 4 Blinkets
3 Blinkets = 7 Drinkets

So, let me express Trinkets in terms of Drinkets.

First, from the second equation: 3 Blinkets = 7 Drinkets => 1 Blinket = 7/3 Drinkets.

Then, substitute into the first equation: 1 Trinket = 4 Blinkets = 4*(7/3 Drinkets) = 28/3 Drinkets.

Therefore, 1 Trinket = 28/3 Drinkets.

Therefore, to find how many Trinkets are in 56 Drinkets, we can divide 56 by (28/3).

Which is 56 * (3/28) = (56/28)*3 = 2*3 = 6.

So, again, 56 Drinkets equals 6 Trinkets.

So, both methods give me the same answer, which is reassuring.

Alternatively, maybe I can use dimensional analysis. Let me write that out.

Starting with 56 Drinkets, and I want to convert to Trinkets.

So, 56 Drinkets * (3 Blinkets / 7 Drinkets) * (1 Trinket / 4 Blinkets).

Let me compute that.

First, the Drinkets unit cancels out: 56 Drinkets * (3 Blinkets / 7 Drinkets) = 56 * 3 / 7 Blinkets.

Then, multiplying by (1 Trinket / 4 Blinkets): (56 * 3 / 7) * (1 / 4) Trinkets.

Calculating the numbers: 56 divided by 7 is 8, 8 times 3 is 24, 24 divided by 4 is 6.

So, again, 6 Trinkets.

So, all methods lead to the same answer. Therefore, I can be confident that 56 Drinkets is equal to 6 Trinkets.

But just to make sure, let me think if there's another way someone might approach this problem and maybe make a mistake. For example, if someone tried to convert Trinkets to Drinkets first, but since we need the reverse, that might complicate things.

Alternatively, someone might miscalculate the conversion factors. For instance, if they thought that 3 Blinkets = 7 Drinkets implies 1 Drinket = 3/7 Blinkets, which is correct, but if they messed up the multiplication or division steps, they might get a wrong answer.

But in my calculations, I checked each step, and they all seem consistent.

Another check: Let's see how many Drinkets are in 1 Trinket.

From above, 1 Trinket = 4 Blinkets. And 3 Blinkets = 7 Drinkets. So, 4 Blinkets = (4/3)*7 Drinkets = 28/3 Drinkets ≈ 9.333 Drinkets.

Therefore, 1 Trinket ≈ 9.333 Drinkets.

Therefore, 56 Drinkets would be 56 divided by (28/3) = 56 * 3 /28 = 6 Trinkets.

Yes, same result.

Alternatively, if I think in terms of ratios.

Let me think of the conversion rates as fractions.

We have:

Trinket to Blinket: 1 T = 4 B => T/B = 4 => B/T = 1/4

Blinket to Drinket: 3 B = 7 D => B/D = 7/3 => D/B = 3/7

Therefore, to get from D to T, we can do D to B to T.

So, D to B is D * (B/D) = D * (7/3) ? Wait, no. Wait, if 3 B = 7 D, then D = (3/7) B. Therefore, to convert D to B, you multiply by (3/7). Wait, no. Wait, if D = (3/7) B, then B = (7/3) D.

Wait, let me clarify.

If 3 B = 7 D, then solving for B gives B = (7/3) D. Therefore, to convert D to B, you multiply by 7/3.

Similarly, to convert B to T, since 1 T = 4 B, then T = B / 4.

Therefore, combining these two steps:

Starting with D, multiply by 7/3 to get B, then divide by 4 to get T.

So, overall, D * (7/3) * (1/4) = D * 7/(12).

Wait, but that would mean that 1 D = 7/12 T.

Wait, but earlier we had 1 T = 28/3 D, so 1 D = 3/28 T.

Wait, there's a discrepancy here. Which is correct?

Wait, let me check.

If 1 T = 4 B, and 3 B = 7 D, then 1 B = 7/3 D. Therefore, 1 T = 4*(7/3 D) = 28/3 D. Therefore, 1 D = 3/28 T.

But according to the other approach, if I convert D to B by multiplying by 7/3, then B to T by multiplying by 1/4, then total factor is 7/3 * 1/4 = 7/12. Therefore, 1 D = 7/12 T?

But that contradicts the previous result.

Wait, so which one is correct?

Wait, there must be a mistake here. Let me check again.

If 3 B = 7 D, then 1 D = 3/7 B.

Therefore, to convert D to B, you multiply by 3/7.

But earlier I thought that B = 7/3 D. Which is correct?

Let me take numbers. Suppose I have 3 B = 7 D.

So, if I have 7 D, that's equal to 3 B. Therefore, 1 D is equal to 3/7 B.

Therefore, to convert D to B, you multiply by 3/7.

But if I have 1 B, that's equal to 7/3 D.

Therefore, if I have 1 D, it's 3/7 B.

Therefore, if I want to convert D to T, I need to first convert D to B, which is multiplying by 3/7, then convert B to T, which is multiplying by 1/4.

Therefore, total factor is 3/7 * 1/4 = 3/28.

Therefore, 1 D = 3/28 T.

Therefore, 56 D = 56 * 3/28 T.

Which is 56 divided by 28 is 2, 2 * 3 = 6.

So, 6 T.

But earlier, when I thought of converting D to B by multiplying by 7/3, that was wrong.

So, where did that mistake come from?

Ah, I think I confused the direction of conversion.

If 3 B = 7 D, then to get B in terms of D, you have B = (7/3) D.

But that would mean that if you have D, multiplying by 7/3 gives you B.

But that would be if you have D and want to get B.

Wait, let me think.

If 3 B = 7 D, then solving for B gives B = (7/3) D.

Therefore, if you have D, multiplying by 7/3 gives you B.

But that would mean that 1 D = 7/3 B?

Wait, no. Wait, if B = (7/3) D, then 1 D = 3/7 B.

Wait, this is confusing. Let me use algebra.

Given 3 B = 7 D.

Solve for B: B = (7/3) D.

Therefore, if you have D, multiplying by 7/3 gives you B.

But that would mean that 1 D = 7/3 B? No, that can't be.

Wait, if B = (7/3) D, then D = (3/7) B.

So, if you have D, you can get B by multiplying D by (7/3).

Wait, let me take numbers. Suppose D = 7. Then, according to the equation, 3 B = 7 D = 7*7 = 49. Therefore, B = 49 / 3 ≈ 16.333.

But according to B = (7/3) D, if D = 7, then B = (7/3)*7 ≈ 16.333. Which matches.

Therefore, if you have D, multiplying by 7/3 gives you B.

But if you have 1 D, then B = 7/3 D.

But that would mean that 1 D is equal to 7/3 B? No, that would be if you had B = 7/3 D, then D = 3/7 B.

So, if you have 1 D, it's equal to 3/7 B.

But if you have 1 B, it's equal to 7/3 D.

Therefore, converting D to B requires multiplying by 3/7, and converting B to D requires multiplying by 7/3.

Therefore, earlier when I thought of converting D to B by multiplying by 7/3, that was wrong. It should be multiplying by 3/7.

So, that was my mistake earlier.

So, going back, when I did the dimensional analysis earlier, I had:

56 Drinkets * (3 Blinkets / 7 Drinkets) * (1 Trinket / 4 Blinkets)

Which is correct because 3 Blinkets / 7 Drinkets is the conversion factor from Drinkets to Blinkets (since 3 Blinkets = 7 Drinkets, so 1 Drinket = 3/7 Blinkets).

Therefore, multiplying by 3/7 gives Blinkets. Then multiplying by 1/4 gives Trinkets.

So, that was correct.

But when I tried to think of converting D to B by multiplying by 7/3, that was wrong. It should be multiplying by 3/7.

So, that was a confusion in the direction of the conversion.

Therefore, the correct conversion factor from D to B is 3/7, not 7/3.

Therefore, the earlier calculation where I had 56 Drinkets * 3/7 Blinkets per Drinket * 1/4 Trinket per Blinket is correct.

So, that gives 56 * 3/7 * 1/4 = 6 Trinkets.

Therefore, the answer is 6 Trinkets.

So, after correcting my earlier confusion, I still arrive at the same answer.

Another way to think about it is to find the exchange rate between Drinkets and Trinkets directly.

We know that 1 Trinket = 4 Blinkets, and 3 Blinkets = 7 Drinkets.

So, let's find how many Drinkets are in 1 Trinket.

Since 1 Trinket = 4 Blinkets, and 3 Blinkets = 7 Drinkets, then 4 Blinkets = (4/3)*7 Drinkets = 28/3 Drinkets.

Therefore, 1 Trinket = 28/3 Drinkets.

Therefore, 1 Drinket = 3/28 Trinkets.

Therefore, 56 Drinkets = 56 * (3/28) Trinkets.

Calculating that: 56 divided by 28 is 2, 2 times 3 is 6.

So, again, 6 Trinkets.

Therefore, regardless of the method, the answer is 6.

I think this is solid. I considered multiple approaches, checked for consistency, corrected my earlier mistake in conversion factors, and all paths lead to the same answer.

Therefore, I can confidently say that 56 Drinkets is equal to 6 Trinkets.

**Final Answer**
The value of 56 Drinkets in Trinkets is \boxed{6}.

To determine the value of **56 Drinkets** in **Trinkets**, we need to use the given conversion relationships:

- **1 Trinket = 4 Blinkets**
- **3 Blinkets = 7 Drinkets**

---

### Step 1: Convert Drinkets to Blinkets

We are given that **3 Blinkets = 7 Drinkets**, so:

$$
1 \text{ Drinket} = \frac{3}{7} \text{ Blinkets}
$$

Now, to find how many Blinkets are in **56 Drinkets**:

$$
56 \text{ Drinkets} \times \frac{3}{7} \text{ Blinkets per Drinket} = \frac{56 \times 3}{7} = \frac{168}{7} = 24 \text{ Blinkets}
$$

---

### Step 2: Convert Blinkets to Trinkets

We are also given that **1 Trinket = 4 Blinkets**, so:

$$
1 \text{ Blinket} = \frac{1}{4} \text{ Trinket}
$$

Now, convert **24 Blinkets** to Trinkets:

$$
24 \text{ Blinkets} \times \frac{1}{4} \text{ Trinket per Blinket} = \frac{24}{4} = 6 \text{ Trinkets}
$$

---

### Final Answer

$$
\boxed{6}
$$

Solution:
Multiply  \[
3\text{ Blinkets}=7\text{ Drinkets}
\] by 8 to find that 24 Blinkets are equivalent to 56 Drinkets.  Multiply  \[
1\text{ Trinkets}=4\text{ Blinkets}
\] by 6 to find that $\boxed{6}$ Trinkets are equivalent to 24 Blinkets (which in turn are equivalent to 56 Drinkets, as we just found).