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Fibonacci series

$F_0=0$, $F_1=1$; $F_{n+1}=F_n+F_{n-1}$

This is a well known identity

$1^2+1^2+2^2+3^2+5^2+\cdots +F_n^2=F_nF_{n+1}$

I was curious and look every websites for a closed form of

$1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=S_n$, but couldn't find one, so I went and look for it.

It takes me a lot of time through experimental mathematics and deduced a closed form for the cube of the Fibonacci series.

$$1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$$

I need someone to verify it by proof. I try next to determine for the 4th power but fail, so I just wonder is there a closed form for nth power Fibonacci series.

Grigory M
  • 17,478

2 Answers2

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For a (somewhat complicated) combinatorial proof see Identity 5 in Benjamin, Carnes, Cloitre. Recounting the Sums of Cubes of Fibonacci Numbers.

Grigory M
  • 17,478
2

This can be proved by induction. It holds for $n=1$. Denoting the right-hand side by $\sigma_n$, we have

\begin{align} \sigma_n+F_{n+1}^3-\sigma_{n+1}&=\frac{F_nF_{n+1}^2+(-1)^{n+1}F_{n-1}+1}2+F_{n+1}^3-\frac{F_{n+1}F_{n+2}^2+(-1)^{n+2}F_n+1}2 \\ &= \frac{2F_{n+1}^3+F_nF_{n+1}^2-F_{n+1}F_{n+2}^2+(-1)^{n+1}F_{n+1}}2 \\ &= \frac{2F_{n+1}^3+F_nF_{n+1}^2-F_{n+1}(F_n+F_{n+1})^2+(-1)^{n+1}F_{n+1}}2 \\ &= \frac{F_{n+1}^3-F_nF_{n+1}^2-F_{n+1}F_n^2+(-1)^{n+1}F_{n+1}}2\;. \end{align}

That this is zero can be confirmed by substituting the explicit form

$$ F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}\;. $$

joriki
  • 238,052