Evaluate $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$ Denote $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$, then how can we prove that

*

*$f(n,1)=\cdots=f(n,n-1)=0,$

*$f(n,n)=\frac{1}{2} (2 n)!,\ \ f(n,n+1)=\frac{1}{12} n (n+1) (2 n+1) (2 n)!, \cdots$
Moreover is there a general closed-form for $f(n,j)$ when $j>n+1$? Any help is appreciated.
 A: I just have one of the answers:
Your formula is:
$$f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$$
Let $j=0$ then it changes to:
$$f(n,0)=\sum _{k=0}^n (-1)^{n-k} \binom{2 n}{n-k}$$
Setting $n-k \mapsto k$ yields:
$$=f(n,0)=\sum _{k=0}^n (-1)^{k} \binom{2 n}{k}=\underbrace{\frac{1}{2}\sum _{k=0}^{2n} (-1)^{k}\binom{2 n}{k}}_\textrm{0}+\frac{1}{2}(-1)^{n}\binom{2 n}{n}$$$$=\color{red}{\frac{1}{2}(-1)^{n}\binom{2 n}{n}}$$
I used the expansion $\left(1-1\right)^{2n}=0$
A: I would like to fill in the details for @MHZ. We seek to evaluate
$$F_{n,j} = \sum_{k=0}^n k^{2j} (-1)^{n-k} {2n\choose n-k}.$$
where $j\ge 1.$ With this in mind we introduce the function
$$F_n(z) = \frac{(2n)!}{2} z^{j-1} \prod_{q=1}^n \frac{1}{z-q^2}.$$
This has the property that the residue at $z=k^2$ where $1\le k \le n$ is
the  desired sum term. We find
$$\mathrm{Res}_{z=k^2} F_n(z) = \frac{(2n)!}{2} k^{2j-2}
\prod_{q=1}^{k-1} \frac{1}{k-q} \frac{1}{k+q}
\prod_{q=k+1}^{n} \frac{1}{k-q} \frac{1}{k+q}
\\ =  \frac{(2n)!}{2} k^{2j-2} \frac{1}{(k-1)!} \frac{k!}{(2k-1)!}
\frac{(-1)^{n-k}}{(n-k)!} \frac{(2k)!}{(n+k)!}
\\ =  \frac{(2n)!}{2} k^{2j-1} \frac{1}{(2k-1)!}
\frac{(-1)^{n-k}}{(n-k)!} \frac{2k (2k-1)!}{(n+k)!}
\\ = (2n)! k^{2j} (-1)^{n-k} \frac{1}{(n-k)!} \frac{1}{(n+k)!}
\\ = \sum_{k=0}^n k^{2j} (-1)^{n-k} {2n\choose n-k}.$$
We now use the fact that residues sum to zero, so our sum becomes
$$- \mathrm{Res}_{z=\infty} F_n(z).$$
Note however that when $n-(j-1) \ge 2$ or $n\ge j+1$ this residue is
zero. Therefore we continue with an Iverson bracket $[[j+1\gt n]]$
(smallest $j$ is $n$)
$$- \mathrm{Res}_{z=\infty} F_n(z)
= \mathrm{Res}_{z=0} \frac{1}{z^2}  F_n(1/z)
\\ = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{j+1}} \prod_{q=1}^n \frac{1}{1/z-q^2}
\\ = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{j+1-n}}
\prod_{q=1}^n \frac{1}{1 - q^2 z}.$$
We thus have immediately by inspection
$$\bbox[5px,border:2px solid #00A000]{
F_{n,n} = \frac{1}{2} (2n)!.}$$
We also get
$$F_{n,n+1} = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{2}}
\prod_{q=1}^{n} \frac{1}{1 - q^2 z}
= \frac{(2n)!}{2} \sum_{q=1}^{n} q^2.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
F_{n,n+1} = (2n)!
\times \frac{1}{12} n (n+1) (2n+1).}$$
Do one more to get
$$F_{n,n+2} = \mathrm{Res}_{z=0}
\frac{(2n)!}{2} \frac{1}{z^{3}}
\prod_{q=1}^{n} \frac{1}{1 - q^2 z}
\\ = \frac{1}{2} (2n)!
\left(\sum_{p=1}^n \sum_{q=p+1}^n p^2 q^2
+ \sum_{q=1}^n q^4\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{
F_{n,n+2} = (2n)!
\times \frac{1}{720} 
n (n+1) (n+2) (2n+1) (2n+3) (5n-1).}$$
