# Prove for any $\sum_{i=1}^n f_i^2 = f_n \times f_{n+1}$ using Fibonacci numbers [duplicate]

I want to prove the below:

$$\sum_{i=1}^n f_i^2 = f_n \times f_{n+1}$$

The example of $$n = 1$$ is trivial:

\begin{align} \sum_{i=1}^1 f_i^2 &= f_1^2 \\ &= 1^2 \\ &= 1 \times 1 \\ &= f_1 \times f_2 \end{align}

And working through $$n=2$$:

\begin{align} \sum_{i=1}^2 f_i^2 &= f_1^2 + f_2^2 \\ &\stackrel{?}{=} f_2 \times f_3 \end{align}

Using the definition of a Fibonacci number, I can work backwards from $$f_2 \times f_3$$:

\begin{align} f_2 \times f_3 &= (f_{2-1}+f_{2-2}) \times (f_{3-1}+f_{3-2}) \\ &= (f_1 + f_0) \times (f_2 + f_1) \\ &= f_1f_2 + f_1f_1 + f_0f_2 + f_0f_1 \\ &= f_1f_2 + f_1^2 \end{align}

Since $$f_2 = 1$$, $$f_1f_2 = f_2f_2 = f_2^2$$, thus $$f_2 \times f_3 = f_1^2 + f_2^2$$.

But how could I generalize this for any $$n$$ and not show examples for $$n=3$$, $$n=4$$, etc?

Use Mathematical Induction. If $$n=1$$, the identity is true. Suppose $$n=k$$ it is true; namely $$\sum_{i=1}^k f_i^2 = f_k \times f_{k+1}.$$ Now for $$n=k+1$$, $$\begin{eqnarray} \sum_{i=1}^{k+1} f_i^2 &=& \sum_{i=1}^k f_i^2+f_{k+1}^2\\ &=&f_k \times f_{k+1}+f_{k+1}^2\\ &=&f_{k+1}(f_k+f_{k+1})\\ &=&f_{k+1}f_{k+2}\\ &=&f_{k+1}f_{(k+1)+1} \end{eqnarray}$$ which implies that the identity is true for $$n=k+1$$. So $$\sum_{i=1}^n f_i^2 = f_n \times f_{n+1}$$ for any $$n$$.