Prove that the Fibonacci numbers obey the identity $\sum_{i=0}^{n} [f(i)]^2=f(n)f(n+1)$ for $n\ge 0$

Question

Prove that the Fibonacci numbers obey the following identity $$\sum_{i=0}^{n} [f(i)]^2=f(n)f(n+1)\;\;\;\;\ n\ge 0$$ Here, $f(0) = 1$.

How can we prove this inequality? I have no idea from here.

Answer 1

An identity of the form $$\sum_{k=0}^n a_k=b_n$$ can always be proved by verifying that $a_0=b_0$ and that $b_{n+1}-b_n=a_{n+1}$.

Here $a_n=f(n)^2$ and $b_n=f(n)f(n+1)$. Then $$b_{n+1}-b_n=f(n+1)f(n+2)-f(n)f(n+1)=f(n+1)[f(n+2)-f(n)]=\cdots$$ etc.