# Fibonacci trick and proving it. [duplicate]

I am trying to learn Fibonacci tricks and I have one that I can not prove. I know it works because Ive tried it multiple times but I have not a clue how to prove. Here it is:

f(0)^2 + f(1)^2 + f(2)^2 + f(3)^2 = f(3)f(3+1)
0    +   1    +   1    +   4    =   2  *  3
= 6                          =6


Is there a way to prove this?

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## marked as duplicate by Martin Sleziak, Davide Giraudo, Najib Idrissi, amWhy, AlexRApr 3 at 13:15

Do you mean to prove that $$f(n)^2+f(n+1)^2+f(n+2)^2+f(n+3)^2=f(n+3)f(n+4)?$$ –  Alex Becker Apr 3 at 4:37
I think he means to prove that $\sum_{i=0}^n f(i)^2=f(n)f(n+1)$. –  Nishant Apr 3 at 4:38
Yes Nishant is correct. –  user081608 Apr 3 at 4:41
Do try induction. One (short) line. –  André Nicolas Apr 3 at 4:46
Im sorry but what do you mean by that Andre? –  user081608 Apr 3 at 4:47

Basis : ok

Inductive step : Assume $∑_{i=0}^{n}f(i)^2=f(n)f(n+1)$. $$∑_{i=0}^{n+1}f(i)^2=f(n)f(n+1)+f(n+1)^2$$$$∑_{i=0}^{n+1}f(i)^2=f(n+1)(f(n)+f(n+1))$$$$∑_{i=0}^{n+1}f(i)^2=f(n+1)f(n+2)$$

There is a geometric interpretation:

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+1 for the geometric interpretation (and of course the correct proof). –  Chris Apr 3 at 11:05

You've already proven that it works for one example $k=3$. Now write up the general equation $$\sum_{k=0}^n F_k^2 = F_n\cdot F_{n+1}$$ and add $F_{n+1}^2$ on both sides. You get $$F_{n+1}^2+ \sum_{k=0}^n F_k^2 =\sum_{k=0}^{n+1} F_k^2 = F_{n+1}^2+ F_n\cdot F_{n+1}=\left(F_{n+1}+ F_n\right)\cdot F_{n+1}$$ and use $F_{n+2}=F_{n+1}+ F_n$, the definition of Fibonacci numbers.

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Since the base case is true, assume that $$\sum_{i=0}^{n}f(i)^2=f(n)f(n+1)$$ is also true and use this assumption to prove that $$\sum_{i=0}^{n+1}f(i)^2=f(n+1)f(n+2).$$

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