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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

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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
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1 Answer 1

up vote 7 down vote accepted

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., $k>2$.

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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
    
Indeed, I just forgot to include the condition on k. Thank you for pointing it out. –  user63566 Feb 23 '13 at 10:17
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