Proving $\sum_{k=0}^{(m-1)/2}(-1)^k{{m+1}\choose{k}}\left(\frac{m+1}{2}-k\right)^p=0$, for odd $m\geq3$ and even $2\leqslant p\leq m-1$

Question

I am trying to prove the following identity for odd $m\geqslant 3$ and even $2\leqslant p\leqslant m-1$:

$$\sum_{k=0}^{(m-1)/2}(-1)^k{{m+1}\choose{k}}\left(\frac{m+1}{2}-k\right)^p=0$$

When I split the sum into two sums, over even $k$ and over odd $k$,

$$\sum_{k=0}^{\frac{m-1}{2}}(-1)^k{{m+1}\choose{k}}\left(\frac{m+1}{2}-k\right)^p=\underbrace{\sum_{k=0}^{\lceil\frac{m+1}{4}\rceil-1}{{m+1}\choose{2k}}\left(\frac{m+1}{2}-2k\right)^p}_{\text{even } k}-\underbrace{\sum_{k=0}^{\lfloor\frac{m+1}{4}\rfloor-1}{{m+1}\choose{2k+1}}\left(\frac{m+1}{2}-2k-1\right)^p}_{\text{odd } k},$$

empirically I get that the two sums are equal. But I have no idea how to prove it.

To make it easier to follow, here are some examples for $p=2$ and $m=3,5,7$:

$${{4}\choose{0}}(2-0)^2={{4}\choose{1}}(2-1)^2$$ $${{6}\choose{0}}(3-0)^2+{{6}\choose{2}}(3-2)^2={{6}\choose{1}}(3-1)^2$$ $${{8}\choose{0}}(4-0)^2+{{8}\choose{2}}(4-2)^2={{8}\choose{1}}(4-1)^2+{{8}\choose{3}}(4-3)^2$$

Any hint is appreciated.

Answer 1

Let $m+1=2n$ and $p=2q$, where $q\in[n-1]$, then the sum is over $k\in[0,n-1]$. Then we need to prove $$ \sum_{k=0}^{n-1}(-1)^k\binom{2n}{k}(n-k)^{2q}=0. $$ Note that $$ \begin{split} \sum_{k=n+1}^{2n}(-1)^k\binom{2n}{k}(n-k)^{2q}&=\sum_{k=0}^{n-1}(-1)^{2n-k}\binom{2n}{2n-k}(n-(2n-k))^{2q}\\ &=\sum_{k=0}^{n-1}(-1)^{2n-k}\binom{2n}{2n-k}(n-(2n-k))^{2q}\\ &=\sum_{k=0}^{n-1}(-1)^{k}\binom{2n}{k}(k-n)^{2q}\\ &=\sum_{k=0}^{n-1}(-1)^{k}\binom{2n}{k}(n-k)^{2q}, \end{split} $$ and $(-1)^n\binom{2n}{n}(n-n)^{2q}=0$, so the desired result is equivalent to $$ \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}(n-k)^{2q}=0, $$ i.e. $$ \sum_{k=0}^{2n}(-1)^{2n-k}\binom{2n}{k}(-n+k)^{2q}=0, $$ But $$ \sum_{k=0}^{2n}(-1)^{2n-k}\binom{2n}{k}(-n+k)^{2q}=(\Delta^{2n}x^{2q}){\large\mid}_{x=-n}, $$ the $2n$-th forward difference of $x^{2q}$ evaluated at $x=-n$ (i.e. $\Delta f(x)=f(x+1)-f(x)$, and $\Delta$ is applied $2n$ times). Note that the application of $\Delta$ decreases the degree of a nonzero polynomial by at least $1$. And since $\deg x^{2q}=2q<2n$, it follows that $\Delta^{2n}x^{2q}=0$ identically as a polynomial. This ends the proof.