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I already know that $f_{n+m}=f_{n-1}f_m + f_nf_{m+1}$.

By letting $m=n$ it immediately follows that $f_{2n}=f_{n}(f_{n+1} + f_{n-1})$ and from that we get $f_{2n}=f^2_{n+1} - f^2_{n-1}$.

From this one should be able to see that $f_{3n}=f^3_{n+1} + f^3_n - f^3_{n-1}$, but I somehow can't figure it out.

Miriam
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  • Have you ever come across this representation of the Fibonacci numbers? I suspect it might provide a neat route to the solution. – πr8 Feb 04 '16 at 17:21
  • Yeah, I know this one. Thanks for bringing it to my attention again. – Miriam Feb 04 '16 at 17:30

1 Answers1

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We have $$\begin{align}f_{n+2n}&=f_{n-1}f_{2n}+f_nf_{n+1+n}\\ &=f_{n-1}(f_{n+1}^2-f_{n-1}^2)+f_n(f_{n}f_{n}+f_{n+1}f_{n+1})\\ &=f_{n-1}f_{n+1}^2-f_{n-1}^3+f_n^3+f_nf_{n+1}^2\\ &=(f_{n-1}+f_n)f_{n+1}^2-f_{n-1}^3+f_n^3\\ &=f_{n+1}^3-f_{n-1}^3+f_n^3\end{align} $$

Hagen von Eitzen
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