<s>[INST] A sum with binomial coefficients Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$
 [/INST] Using these three identities for $n^{\text{th}}$ differences:
$$
\begin{align}
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^n&=n!\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+1}&=n!\binom{n+1}{2}\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+2}&=n!\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)
\end{align}
$$
and the fact that the $n^{\text{th}}$ difference of a polynomial of degree less than $n$ is $0$, we get
$$
\begin{align}
&\sum_{k=0}^n(-1)^k\binom{n}{k}(n-2k)^{n+2}\\
&=2^{n+2}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(k-n/2)^{n+2}\\
&=2^{n+2}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\left(k^{n+2}-\frac{(n+2)n}{2}k^{n+1}+\frac{(n+2)(n+1)n^2}{8}k^n+\dots\right)\\
&=2^{n+2}n!\left(\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)-\frac{(n+2)n}{2}\binom{n+1}{2}+\frac{(n+2)(n+1)n^2}{8}\right)\\
&=2^nn!\binom{n+2}{3}\\[6pt]
&=\frac{2^nn(n+2)!}{6}
\end{align}
$$
</s>