Help with a proof I can't quite

Question

Let $(F_j)^\infty_{j=1}$ be the sequence of Fibonacci numbers. For all $n \in \mathbb{N}$, $\sum\limits_{k=1}^{2n-1}F_kF_{k+1}=(F_{2n})^2$.

I handled the base case quite well but couldn't go very far on the induction step. Any help proving would be quite helpful.

Answer 1

Suppose it's true for $n$. We want to show it is true for $n+1$ so we need to examine

$$\sum_{k=1}^{2n+1}F_kF_{k+1}.$$

But breaking off the last two terms we see that this is nothing more than

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

We want to show that this is equal to $(F_{2n+2})^2$. We know that $F_{2n+1} = F_{2n+2}-F_{2n}$ since $(F_n)_{n=1}^{\infty}$ is the Fibonacci sequence. Can you take it from here?

Answer 2

You know that $F_{k+1}=F_k+F_{k-1}$ and perhaps also that $F_{k-1}F_{k+1}-F_k^2=(-1)^k$. Then $$ F_kF_{k+1}=(F_{k+1}-F_{k-1})F_{k+1}=F_{k+1}^2-(F_k^2+(-1)^k)=F_{k+1}^2-F_k^2-(-1)^k $$ so that your sum transforms in a telescoping and an oscillating part, $$ \sum_{k=1}^NF_kF_{k+1}=F_{N+1}^2-F_1^2-\sum_{k=1}^N(-1)^k=F_{N+1}^2-\tfrac12(1+(-1)^N) $$