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This question already has an answer here:

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, Namaste, AlexR Apr 3 '14 at 13:15

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  • $\begingroup$ Do you mean to prove that $$f(n)^2+f(n+1)^2+f(n+2)^2+f(n+3)^2=f(n+3)f(n+4)?$$ $\endgroup$ – Alex Becker Apr 3 '14 at 4:37
  • $\begingroup$ I think he means to prove that $\sum_{i=0}^n f(i)^2=f(n)f(n+1)$. $\endgroup$ – Nishant Apr 3 '14 at 4:38
  • $\begingroup$ Yes Nishant is correct. $\endgroup$ – user081608 Apr 3 '14 at 4:41
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    $\begingroup$ Do try induction. One (short) line. $\endgroup$ – André Nicolas Apr 3 '14 at 4:46
  • $\begingroup$ Im sorry but what do you mean by that Andre? $\endgroup$ – user081608 Apr 3 '14 at 4:47
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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:

a

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    $\begingroup$ +1 for the geometric interpretation (and of course the correct proof). $\endgroup$ – Chris Apr 3 '14 at 11:05
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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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