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

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 $$F_{2k}$$?

• Note the Fibonacci number Wikipedia entry states that a form of "d'Ocagne's identity" is $F_{2n} = F_{n + 1}^2 - F_{n - 1}^2$. Thus, doing a search on this identity may provide a relatively simple way to prove it and, thus, finish your induction proof. Commented Jan 23, 2019 at 2:15
• – Sil
Commented Jun 26, 2021 at 20:30

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 $$F_{n+1}F_{n}-F_{n-1}F_{n-2}=F_{2n-1}$$.
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.