<s>[INST] A sum of Fibonacci numbers Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity: 
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k}$$
How can I evaluate the following sums to show this identity (1) is correct?
$$\sum_{n=0}^{\infty}(-1)^n\frac{n^2}{{2n \choose n}(2n+1)}=A$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{{2n \choose n}(2n+1)}=B$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{1}{{2n \choose n}(2n+1)}=C$$
 [/INST] Here is another answer, in the same spirit as that of @dan_fulea:
Both sides of $$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k+1}$$ can be considered as sequences defined by the same second order linear recurrence for the index $k$. Then, for them to be equal for all $k$, it suffices that they coincide at the first values $0$ and $1$ of the index $k$.
In other words, we just need to show that 
\begin{align}\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&= F_1=1
\end{align} 
and
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n^2+9n+4}{2n+1}\frac{1}{{2n \choose n}}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n} &=F_2=1
\end{align}
where we have introduced the Catalan number $C_n=\frac{1}{n+1} \binom{2n}{n}$.
But it is easy to verify, by induction, that 
\begin{align}\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+3}{2n+1}\frac{1}{{2n \choose n}}&=\frac{{2m+2 \choose m+1}+(-1)^m}{{2m+2 \choose m+1}}\\
\sum_{n=0}^{m}\frac{(-1)^n}{2}\frac{5n+4}{2n+1}\frac{1}{C_n}&=\frac{C_{m+1}+(-1)^m}{C_{m+1}}
\end{align} 
and we are done.
</s>