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I would like to calculate, or bound from above, the following sum $$ \sum_{i=0}^n(n-2i)^p{p \choose i}, $$ here $p\geq 2$.

Any references are very welcome.

Thank you.

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Well, it is not bigger than $(2 n)^p$ – Igor Rivin Apr 13 '12 at 16:35
$p$ fixed, $n\to\infty$? $n$ fixed, $p\to\infty$? $n,p$ both going to infinity in some unspecified way? – Gerry Myerson Apr 13 '12 at 23:14
The paper has some results that seem quite similar, so perhaps the methods used there can be adapted to Michael's problem. – Richard Stanley Apr 28 '12 at 21:30
But why crosspost under different names? Or am I missing something? – Felix Goldberg Sep 18 '12 at 13:06
Not relevant to your question, but it can be shown that all the zeros of the polynomial $\sum_{i=0}^p (x-2i)^p{p\choose i}$ have real part $p$. – Richard Stanley Oct 24 '13 at 0:17

On of methods for finding an approximation or bound for a finite sum is using the following formula $$ \sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$$ and by taking $f(i)= (n-2i)^p\binom {p}{i}$ you can consider decimal digits of right hand side which are faster than of left hand side. Also $B_k$ here are Bernoulli numbers.

Moreover, the sharp bounds of Bernoulli numbers has been computed ,(see here ) So you can also try to find a bound for your sum.

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