I came across the following alternating sum
$$
\sum\_{k=0}^n (-1)^k \binom{2n}{k} (n-k)^r,\quad 1\leq r < n.
$$
It seems that when $r$ is an even integer the sum is $0$ and when $r$ is an odd integer the sum is not zero (regardless of the parity of $n$). 


[Edited] The case when $r$ is even is easy by symmetry as Darij Grinberg pointed out below. So the question left is how to show that the sum is nonzero when $r$ is odd.


The main difficulty I have proving this is that the sum is only from $0$ to $n$ instead of to $2n$, and I don't see how to apply the classical methods such as finite differences, integral representation, series, etc.



partial alternating sum involving binomial coefficients
In [a post on Math Stack Exchange, MSE 2827591](https://math.stackexchange.com/questions/2827591/alternate-proof-for-weighted-alternating-shifted-central-binomial-sum-relation), I prove the following:


$$\sum\_{k=1}^n (-1)^{k+1} \binom{2n}{n+k} k^s =
\binom{2n}{n} \sin(\pi s/2) \int\_0^\infty \frac{dx \, \,x^s}{\sinh{\pi x}} \frac{n!^2}{(n+ix)!(n-ix)!}.
$$ 


The OP's formula can be put in the form on the LHS of this equation. By inspection the questions of concern can be answered; namely, the value of zero for $s$ an even integer, and non-zero otherwise.


