<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] 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
$$
F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}\;.
$$
</s>