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I'm working on a problem that asks us to prove for any n $\geq$ 1, $F_n$ the Fibonacci number satisfy $F_{n+1}^2 - F_{n-1}^2 = F_{2n}$.

I learned Fibonacci identity $F_{2n} = F_{1}+F_{3}+F_{5}+...+F_{2n-1}$ by combinatorial-proofs but I don't know how to deal with the LHS.

Any help would be appreciated.

zooooo
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2 Answers2

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

For a proof of the first equality, see this answer.

lhf
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Let $$\mathcal{F} = \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix} $$ and $$\mathcal{F}_n = \begin{pmatrix} F_{n+1}\\ F_n \end{pmatrix} $$

we have $$\mathcal{F}_{n+1} = \mathcal{F}\mathcal{F}_n$$ so $$\mathcal{F}^n = \begin{pmatrix} F_{n+1} & F_n\\ F_n & F_{n-1} \end{pmatrix} $$

now write $\mathcal{F}^{2n} = \mathcal{F}^n \mathcal{F}^n$ we deduce among other identities \begin{align} F_{2n} &= F_nF_{n+1} + F_nF_{n_1}\\ &= F_n\left(F_{n+1} + F_{n-1}\right)\\ &= \left(F_{n+1} - F_{n-1}\right)\left(F_{n+1} + F_{n-1}\right)\\ &= F_{n+1}^2 - F_{n-1}^2 \end{align}

user388615
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