Prove $F_n=(-1)^{n+1}F_{-n}$ without induction

Question

Consider the Fibonacci sequence defined by $$F_{n+2}=F_{n+1}+F_n,~~~~~~F_1=F_2=1$$ I can prove via induction that $F_n=(-1)^{n+1}F_{-n}$, but how can it be proven without using induction?

Thank you for your help.

Answer 1

Binets formula $$F_n = \frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}},$$ where $\phi$ denotes the golden ratio, is an explicit formula for the $n$-th Fibonacci number $F_n$. We have to prove that $$\underbrace{\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}}_{F_n}=(-1)^{n+1}\cdot\underbrace{\frac{\phi^{-n}-(-\phi)^{n}}{\sqrt{5}}}_{F_{-n}}.$$ We can rewrite $(-1)^{n+1}\phi^{-n}$ to $-(-\phi)^{n}.$ Also $(-1)^{n+1}(-\phi)^{-n}$ is equal to $-\phi^n$. Putting all of this together yields $$(-1)^{n+1}\cdot{\frac{\phi^{-n}-(-\phi)^{n}}{\sqrt{5}}} = \frac{-(-\phi)^{-n}-[-\phi^{n}]}{\sqrt{5}}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} = F_n$$

Answer 2

Instead define $F_{-k}:=(-1)^{k+1}F_k$ for $k>0$, then verify the recursion relation still works viz.$$F_{-k}-F_{-k-1}-F_{-k-2}=(-1)^{k+1}(F_k+F_{k+1}-F_{k+2})=0.$$