Closed-form expression for $\sum_{n=1}^{k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}$? Wolframalpha tells me that
$$\sum_{n=1}^{k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}=0$$ for $k>2$
However I have not been able to come up with a proof and I simply don't see how to do it. Does anyone have the slightest idea?
Anything is welcome: Proofs, hints, or references.
Thank you very much in advance
 A: We have
\begin{eqnarray*}
&&\sum_{1\le n\le k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}\\
&=&\frac12 \sum_{1\le n\le k} (-1)^{n+1}n^2(n^2-1)\left(\binom{2k}{k-n}+\binom{2k}{k+n}\right),\\
&& \qquad \ \ \ \text{since } \binom{2k}{k-n} = \binom{2k}{k+n} \\
&=&\frac12 \sum_{-k\le n\le k, \ n\ne 0} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k+n},\\
&& \qquad \ \ \ \text{rearranging terms and reindexing}\\
&=&\frac12 \sum_{-k\le n\le k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k+n}\\
&=&\frac12 (-1)^{k+1} \sum_{0\le \ell\le 2k} (-1)^{\ell} (\ell-k)^2((\ell-k)^2-1)\binom{2k}{\ell},\\
&& \qquad \ \ \ \text{reindexing using } \ell=k+n.
\end{eqnarray*}
Now, for any polynomial $P(M,\ell)$ of degree $d$ in $\ell$, the sum
$$
\sum_{0\le \ell\le M} (-1)^{\ell} P(M,\ell)  \binom{M}{\ell}
$$
will vanish if $M$ exceeds $d$.  The reason is that $P$ can be expressed as a linear combination of the first $d+1$ of the basis polynomials
$$1, \ \ \ell,\ \ \binom{\ell}{2}, \ \ \dots, \binom{\ell}{i}, \dots$$
with coefficients which are polynomials in $M$, and then, for each $j=0$, $\dots$, $d$,
\begin{eqnarray*}
&&\sum_{0\le \ell\le M} (-1)^{\ell} \binom{\ell}{j} \binom{M}{\ell}\\
&=&\binom{M}{j} \sum_{j\le \ell\le M} (-1)^{\ell} \binom{M-j}{\ell-j}\\
&=& \binom{M}{j} (-1)^j (1-1)^{M-j}\\
&=& 0.
\end{eqnarray*}
In the case here, we have $M=2k$, $P(M,\ell)=(\ell-\frac M2)^2((\ell-\frac M2)^2-1)$, and $d=4$, so the sum will vanish whenever $2k>4$, i.e., $k>2$.
