# computation of the sum

I am having trouble to compute the following sum: $$\sum_{k=0}^n(n-2k)^p \frac{{n \choose k}{2m-n \choose m-k}}{{2m \choose m}}$$ Here $p\geq 2$.

To simplify the question, we can even assume that $n/2-C\sqrt n\leq k \leq n/2+C\sqrt n$. Any help or sources will be very helpful.

Thank you.

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So you asymptotics or exact value? –  no identity Apr 11 '12 at 23:52
I would like to get exact value, but I don't know even how to get asymptotic. I will apreciate any explanation. –  Michael Apr 12 '12 at 0:58
Even Mathematica doesn't know the answer... –  no identity Apr 12 '12 at 9:41
I guess it's implied that $n\le p$. Besides, the first factor can be negative (if p is odd). Nothing wrong with that, but feels a little strange. Are you sure you got it right? –  leonbloy Apr 12 '12 at 15:07
Yes. Its right. What if to make substitution $2k-n=m$? –  Michael Apr 12 '12 at 15:35

The search term "moment generating function of hypergeometric distribution" may produce some information. One form of the MGF is listed at the Wikipedia page on the hypergeometric distribution, but probably you know that and want a form where the symmetry around $n/2$ is used. Posting to stats.stackexchange.com or adding a (statistics) tag could be useful.

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The expression you gave can be simplified to

$$S = \sum_{k=0}^n(n-2k)^p \frac{{m \choose {n-k}}{m \choose k}}{{2m \choose n}}$$

This is same as

$$S = \sum_{k=0}^n(2k-n)^p \frac{{m \choose {n-k}}{m \choose k}}{{2m \choose n}}$$

by substituting n-k as k.

So at least in the case where p is odd, this shows that S = -S, so S = 0; not sure what to do when p is even.

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