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Recall the Fibonacci function defined by $f(0) = 0; \\f(1) = 1; \\f(n) = f(n-1) + f(n -􀀀 2)$

for all $n \ge 2$ Prove that $f(n) \cdot f(n + 1) = f(1)^2 + f(2)^2 + . . . + f(n)^2.$

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    $\begingroup$ Try using induction on $n$. $\endgroup$ – rogerl Feb 11 '16 at 21:07
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    $\begingroup$ What have you tried? Yes, induction is the way to go. Can you do the base case? This answer is a great writeup $\endgroup$ – Ross Millikan Feb 11 '16 at 21:07
  • $\begingroup$ yes i did the base case and after progressing i got this : 3.[ f(1)^2+f(2)^2+..+f(k)^2+ f(k-1)^2]+2.f(k)^2 which is not the desired formula $\endgroup$ – user312759 Feb 11 '16 at 21:08
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Hint: $$f(n)\cdot f(n+1) = \sum_{k=1}^nf(k)^2$$ $$\implies f(n+1)\cdot f(n+2) = f(n+1)\left[f(n+1) + f(n)\right]$$ $$= f(n+1)^2 + f(n)f(n+1) = \cdots$$

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