To prove the given statement by induction, let’s tackle it in two parts. First, we’ll show that $f_{n+m} = f_{n-1} f_m + f_n f_{m+1}$ for all $m \in \mathbb{N}$ and $n \geq 2$. Then we’ll demonstrate that this implies the specific relation $f_{2n} = f_n (f_{n-1} + f_{n+1}) = f_{n+1}^2 - f_{n-1}^2$ for all $n \geq 2$.

Part (a): Inductive Proof of $f_{n+m} = f_{n-1} f_m + f_n f_{m+1}$

1. Base Case: Let's prove the base case where $m = 1$.

   • We need to show that $f_{n+1} = f_{n-1} f_1 + f_n f_2$.
   • Since $f_1 = f_2 = 1$, we have: $f_{n-1} f_1 + f_n f_2 = f_{n-1} \cdot 1 + f_n \cdot 1 = f_{n-1} + f_n.$
   • By the Fibonacci definition, $f_{n+1} = f_{n-1} + f_n$, so the base case holds.

2. Inductive Hypothesis: Assume that for some $m \in \mathbb{N}$, the formula holds: $f_{n+m} = f_{n-1} f_m + f_n f_{m+1}.$

3. Inductive Step: We need to prove the statement for $m + 1$: $f_{n + (m + 1)} = f_{n-1} f_{m+1} + f_n f_{m+2}.$ Using the Fibonacci recurrence relation, we know: $f_{n + (m+1)} = f_{n+m+1} = f_{n+m} + f_{n+m-1}.$ By the inductive hypothesis, we can substitute $f_{n+m}$ and $f_{n+m-1}$: $f_{n+m} = f_{n-1} f_m + f_n f_{m+1},$ $f_{n+m-1} = f_{n-1} f_{m-1} + f_n f_m.$ Therefore, $f_{n+m+1} = (f_{n-1} f_m + f_n f_{m+1}) + (f_{n-1} f_{m-1} + f_n f_m).$ Rearranging terms, we have: $f_{n+m+1} = f_{n-1} (f_m + f_{m-1}) + f_n (f_{m+1} + f_m).$ Using the Fibonacci property $f_{m+1} = f_m + f_{m-1}$ and $f_{m+2} = f_{m+1} + f_m$, we find: $f_{n+m+1} = f_{n-1} f_{m+1} + f_n f_{m+2}.$ This completes the inductive step, so by induction, the formula holds for all $m \in \mathbb{N}$ and $n \geq 2$.

Part (b): Proving $f_{2n} = f_n (f_{n-1} + f_{n+1}) = f_{n+1}^2 - f_{n-1}^2$

Using the result from Part (a) for $m = n$, we have: $f_{2n} = f_{n-1} f_n + f_n f_{n+1} = f_n (f_{n-1} + f_{n+1}).$

To show the equivalence with $f_{2n} = f_{n+1}^2 - f_{n-1}^2$:

1. Start with the identity: $f_{n+1}^2 - f_{n-1}^2 = (f_{n+1} - f_{n-1})(f_{n+1} + f_{n-1}).$
2. Using the Fibonacci recurrence, we know $f_{n+1} - f_{n-1} = f_n$. Substitute this into the expression: $f_{n+1}^2 - f_{n-1}^2 = f_n (f_{n+1} + f_{n-1}).$ Thus, we have shown: $f_{2n} = f_n (f_{n-1} + f_{n+1}) = f_{n+1}^2 - f_{n-1}^2,$ which completes the proof.

Summary

This concludes the inductive proof of both parts. Let me know if you need more details or clarification on any step!

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Related Questions:

1. What is the next step if we want to generalize this formula further?
2. How can we use matrix representations to prove properties of Fibonacci numbers?
3. Can we extend this formula to other sequences defined by a similar recurrence relation?
4. How does this identity help in computational applications involving Fibonacci numbers?
5. Can you find a similar identity for the Lucas sequence?

Tip:

For complex Fibonacci proofs, breaking down the recurrence into smaller cases often simplifies the process.