Evaluate $f(n,j)=\sum _{k=0}^n k^{2 j} (-1)^{n-k} \binom{2 n}{n-k}$

Question

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.

Answer 1

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

$$\bbox[5px,border:2px solid #00A000]{ F_{n,n+2} = (2n)! \times \frac{1}{720} n (n+1) (n+2) (2n+1) (2n+3) (5n-1).}$$

Answer 2

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 $\left(1-1\right)^{2n}=0$