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Let $(F_n)_{n\in \mathbb{N}}$ be the Fibonacci sequence. Prove that $F_{n+1}F_n - F_{n-1}F_{n-2} = F_{2n-1}$.

I'm trying to prove usinge the induction principle, so here is my sketch:

$(i)$ $n = 3 \implies F_4F_3-F_2F_1= 6-1 = 5 = F_5$

$(ii)$ Supose true for $n = k$

$F_{k+2}F_{k+1}-F_{k}F_{k-1} = (F_{k+1}+F_k)F_{k+1} - (F_{k-1}+F_{k-2})F_{k-1} = F_{k+1}F_k-F_{k-1}F_{k-2} + F_{k+1}^2 - F_{k-1}^2 = F_{2k-1} + F_{k+1}^2 - F_{k-1}^2 $.

I got stuck here, how can I transform $F_{k+1}^2 - F_{k-1}^2$ in $F_{2k}$?

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2 Answers 2

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First we can prove that $F_n=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n)$ by induction on $n$. Using the above formula, we can verify that $F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$.

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I think you should use the recurrence formula of the Fibonnacci sequence, I.e $ F(n+2) = F(n+1) + F(n) $ Combine this with the formula you get by induction.

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