Work out the value of Fn−1Fn+1 −Fn2 for n = 1,2,3,4,5. You should see a pattern; prove that this pattern always holds.

Let's begin by calculating the expression \( F_{n-1}F_{n+1} - F_n^2 \) for the first few values of \( n \), where \( F_n \) denotes the \( n \)-th Fibonacci number.

### Step 1: Recall the Fibonacci sequence
The Fibonacci sequence is defined as:
\[
F_1 = 1, \quad F_2 = 1
\]
\[
F_n = F_{n-1} + F_{n-2} \quad \text{for } n \geq 3
\]
So, the first few Fibonacci numbers are:
\[
F_1 = 1, \quad F_2 = 1, \quad F_3 = 2, \quad F_4 = 3, \quad F_5 = 5, \quad F_6 = 8, \ldots
\]

### Step 2: Calculate the expression for \( n = 1, 2, 3, 4, 5 \)

1. **For \( n = 1 \):**
   \[
   F_0 = 0, \quad F_2 = 1, \quad F_1 = 1
   \]
   \[
   F_0F_2 - F_1^2 = 0 \times 1 - 1^2 = 0 - 1 = -1
   \]

2. **For \( n = 2 \):**
   \[
   F_1 = 1, \quad F_3 = 2, \quad F_2 = 1
   \]
   \[
   F_1F_3 - F_2^2 = 1 \times 2 - 1^2 = 2 - 1 = 1
   \]

3. **For \( n = 3 \):**
   \[
   F_2 = 1, \quad F_4 = 3, \quad F_3 = 2
   \]
   \[
   F_2F_4 - F_3^2 = 1 \times 3 - 2^2 = 3 - 4 = -1
   \]

4. **For \( n = 4 \):**
   \[
   F_3 = 2, \quad F_5 = 5, \quad F_4 = 3
   \]
   \[
   F_3F_5 - F_4^2 = 2 \times 5 - 3^2 = 10 - 9 = 1
   \]

5. **For \( n = 5 \):**
   \[
   F_4 = 3, \quad F_6 = 8, \quad F_5 = 5
   \]
   \[
   F_4F_6 - F_5^2 = 3 \times 8 - 5^2 = 24 - 25 = -1
   \]

### Step 3: Observe the pattern
From the calculations above, we notice the following pattern:
- When \( n \) is odd, \( F_{n-1}F_{n+1} - F_n^2 = -1 \).
- When \( n \) is even, \( F_{n-1}F_{n+1} - F_n^2 = 1 \).

### Step 4: Prove that this pattern always holds

We need to prove that for any \( n \geq 1 \):
\[
F_{n-1}F_{n+1} - F_n^2 = (-1)^n
\]

#### Proof by Induction

**Base Case:**  
For \( n = 1 \), we've already calculated:
\[
F_{0}F_{2} - F_{1}^2 = 0 \times 1 - 1^2 = -1
\]
which is \( (-1)^1 \).

**Inductive Step:**  
Assume that for some \( n = k \), the pattern holds:
\[
F_{k-1}F_{k+1} - F_k^2 = (-1)^k
\]
We need to show that it holds for \( n = k + 1 \), i.e., we need to prove:
\[
F_{k}F_{k+2} - F_{k+1}^2 = (-1)^{k+1}
\]

Using the Fibonacci identity \( F_{k+2} = F_{k+1} + F_k \), we can rewrite the left side:
\[
F_k F_{k+2} - F_{k+1}^2 = F_k (F_{k+1} + F_k) - F_{k+1}^2
\]
\[
= F_k F_{k+1} + F_k^2 - F_{k+1}^2
\]
\[
= (F_k F_{k+1} - F_{k+1}^2) + F_k^2
\]
\[
= F_{k+1}(F_k - F_{k+1}) + F_k^2
\]
But since \( F_k - F_{k+1} = -F_{k-1} \), this simplifies to:
\[
= -F_{k+1}F_{k-1} + F_k^2
\]
\[
= -(F_{k-1}F_{k+1} - F_k^2)
\]
By the inductive hypothesis:
\[
= -((-1)^k) = (-1)^{k+1}
\]

Thus, the pattern \( F_{n-1}F_{n+1} - F_n^2 = (-1)^n \) holds for all \( n \geq 1 \).

Would you like to explore more about this proof or Fibonacci identities?

### Related Questions:
1. How can you prove other identities involving Fibonacci numbers?
2. What is the significance of the alternating pattern in Fibonacci-related expressions?
3. Can this pattern be generalized to other recursive sequences?
4. How does the Fibonacci sequence relate to the golden ratio?
5. What are the applications of Fibonacci identities in real-world problems?

### Tip:
When dealing with sequences like Fibonacci, always try to use induction or known identities to simplify complex expressions.

Explore the Fibonacci sequence pattern Fn−1Fn+1 − Fn^2 through detailed calculations and proofs. This mathematical problem demonstrates the application of Fibonacci identities and mathematical induction. Suitable for advanced high school students and anyone interested in recursive sequences.

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