<s>[INST] Proof of $1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$ 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.
 [/INST] For a (somewhat complicated) combinatorial proof see Identity 5 in Benjamin, Carnes, Cloitre. Recounting the Sums of Cubes of Fibonacci
Numbers.
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