Prove $F_{1}^{2}+F_{2}^{2}+\dots+F_{n}^{2}=F_{n}F_{n+1}$ using geometric approach

Question

Prove the identity: $F_{1}^{2}+F_{2}^{2}+\dots+F_{n}^{2}=F_{n}F_{n+1}$, where $F_i$ denotes a Fibonacci number.

How can I prove it using a geometric approach?

Proof by induction is next-to-trivial. Why would you desire a geometric approach and how would you geometrically interpret $F_i$? — AlexR, Aug 06 '13 at 17:07
possible duplicate of For the Fibonacci numbers, show for all $n$: $F_1^2+F_2^2+\dots+F_n^2=F_nF_{n+1}$ — Cameron Buie, Sep 29 '13 at 21:39

A.E · Answer 1 · 2013-08-06T17:13:37.220

16

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The horizontal side length is $F_{n+1}$, in this case $21 + 13=34$. The vertical side is $F_n$, in this case $13 + 8 = 21$. However, the area can also be defined as the sum of the smaller squares.

edited Aug 06 '13 at 17:13

answered Aug 06 '13 at 17:08

A.E

2,473

sounds good appreciatted! – user89260 Aug 06 '13 at 17:14
That is too cool! – Robert Lewis Sep 05 '13 at 19:26

Prove $F_{1}^{2}+F_{2}^{2}+\dots+F_{n}^{2}=F_{n}F_{n+1}$ using geometric approach

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