Recently I came across a finite sum which appears to be zero for all odd numbers. The sum is defined as follows:
$$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)(-1)^k$$ where $n=2m+1$.
For the first few $m$ this sum always equals zero. I tried to prove this by induction with the inductive step from $m=k$ to $m=k+1$ but I did not got this far with this approach. So I am asking for a proof of this formula with an explanation.
I have found more of these sums and I would be interested in a more general result. It seems to me like that in general the exponent of the term $n-2k$ is irrelevant as long as it is an odd number. So that
$$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^3(-1)^k$$ $$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^5(-1)^k$$ $$...$$
all equal zero. I guess thats a fact but I have no clue how to derive this.
I have to add that it is not needed for the number to be odd or even - all these sums should work out for both. So also the following should equal zero.
$$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^2(-1)^k$$ $$\sum_{k=0}^{m}~\binom{n}{k}(n-2k)^4(-1)^k$$ $$...$$