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

• Have you tried a proof by induction? – CogitoErgoCogitoSum Feb 22 '13 at 18:40
• Well, the basis $k=3$ is true :) $\sum_{n=1}^{3} (-1)^{n+1}n^2(n^2-1)\binom{2*3}{3-n} = 0 + -72 + 72 = 0$ – user17753 Feb 22 '13 at 21:37

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., $k>2$.
• Thank you very much for your answer. Indeed using your results, I think the identity can even be generalised to $\sum_{n=1}^{k}(-1)^{n+i+1}n^2(n^2-1)(n^2-4)\cdots(n^2-i^2)\binom{2k}{k-n}=0$. – user63566 Feb 23 '13 at 9:28
• Yes, although you then need $k>i+1$. – David Moews Feb 23 '13 at 9:39