I have numerical evidence that
$$ \sum_{k=1}^n (-1)^k\frac{k^{p}}{n+k}\binom{2n-1}{n-k}=0 $$
For $p=2,4,6...2n-2$.
How could this be proved?
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This sum can be expressed in terms of $2n$-th forward difference: \begin{split} \sum_{k=1}^n (-1)^k\frac{k^{p}}{n+k}\binom{2n-1}{n-k} &= \frac1{2n} \sum_{k=1}^n (-1)^k k^{p} \binom{2n}{n+k} \\ &\mathop{=}_{p\ \text{even}}\frac{(-1)^n}{4n} \sum_{k=0}^{2n} (-1)^k (k-n)^{p} \binom{2n}{k} \\ &=\frac{(-1)^n}{4n}\left.\Delta^{2n}x^p\right|_{x=-n}, \end{split} which is zero for $p<2n$.
answered Feb 28, 2022 at 14:38
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$\begingroup$ Thank you. How do you go from the second step to the third, where the range goes from $0$ to $2n$? $\endgroup$ Commented Mar 2, 2022 at 5:27
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1$\begingroup$ By symmetry of binomial coefficients and change of summation index $k\to k-n$. $\endgroup$ Commented Mar 2, 2022 at 12:36