<s>[INST] Evaluate $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$ Denote $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$, then how can we show that:

*

*$g(n,1)=\frac{n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}$

*$g(n,n)=\frac{\left(\frac{1}{2} n (n+1)\right)^{n-1} n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}$

*$g(n,n+1)=\frac{\left(\frac{1}{2} n (n+1)\right)^n n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}-n!$
This post is related. Any help will be appreciated!
 [/INST] We seek to evaluate
$$G_{n,j} = \sum_{k=1}^n \frac{k^j (-1)^{n-k} {n\choose k}} 
{\frac{1}{2}n(n+1)-k}.$$
With this in mind we introduce the function
$$F_n(z) = n! \frac{z^{j-1}}{\frac{1}{2}n(n+1)-z}
\prod_{q=1}^n \frac{1}{z-q}.$$
This has the property that the residue at $z=k$ where $1\le k\le n$ is
the desired sum  term. We find
$$\mathrm{Res}_{z=k} F_n(z) 
= n! \frac{k^{j-1}}{\frac{1}{2}n(n+1)-k}
\prod_{q=1}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^{n} \frac{1}{k-q}
\\ = n! \frac{k^{j}}{\frac{1}{2}n(n+1)-k}
\frac{1}{k} \frac{1}{(k-1)!}
\frac{(-1)^{n-k}}{(n-k)!}
\\ = \frac{k^{j}}{\frac{1}{2}n(n+1)-k}
(-1)^{n-k} {n\choose k}.$$
We will evaluate this using the fact that residues sum to zero and if
$(n+1)-(j-1) \ge 2$ or $n\ge j$ the residue at infinity is zero, so we 
have in this case
$$G_{n,j} = - \mathrm{Res}_{z=\frac{1}{2} n(n+1)} F_n(z)
= n! \frac{(\frac{1}{2} n(n+1))^{j-1}}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}.$$
We thus have
$$\bbox[5px,border:2px solid #00A000]{
G_{n,1} = \frac{n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}}$$
and
$$\bbox[5px,border:2px solid #00A000]{
G_{n, n} = \frac{(\frac{1}{2} n(n+1))^{n-1} n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}.}$$
When $j\gt n$ we must use the formula
$$G_{n,j} = - \mathrm{Res}_{z=\frac{1}{2} n(n+1)} F_n(z)
- \mathrm{Res}_{z=\infty} F_n(z).$$
We  have
$$- \mathrm{Res}_{z=\infty} F_n(z)
= \mathrm{Res}_{z=0} \frac{1}{z^2} F_n(1/z)
\\ = n! \times \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1}{z^{j-1}}
\frac{1}{\frac{1}{2}n(n+1)-1/z}
\prod_{q=1}^n \frac{1}{1/z-q}
\\ = n! \times \mathrm{Res}_{z=0} \frac{1}{z^{j+1}}
\frac{z}{\frac{1}{2}n(n+1)z-1}
\prod_{q=1}^n \frac{z}{1-qz}
\\ = n! \times \mathrm{Res}_{z=0} \frac{1}{z^{j-n}}
\frac{1}{\frac{1}{2}n(n+1)z-1}
\prod_{q=1}^n \frac{1}{1-qz}.$$
In particular when $j=n+1$ we just need the constant term and find
$$n! \frac{1}{\frac{1}{2}n(n+1)\times 0 -1}
\prod_{q=1}^n \frac{1}{1-q\times 0} = -n!$$
we thus have
$$\bbox[5px,border:2px solid #00A000]{
G_{n, n+1} = \frac{(\frac{1}{2} n(n+1))^{n} n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)} - n!.}$$
The general case for $j\gt n$ is
$$n! \times \mathrm{Res}_{z=0} \frac{1}{z^{j}}
\frac{1}{\frac{1}{2}n(n+1)z-1}
\prod_{q=1}^n \frac{z}{1-qz}$$
which yields
$$-n! \sum_{q=0}^{j-1} \left(\frac{1}{2}n(n+1)\right)^q
{j-1-q\brace n}$$
so that the closed form is (here we must have $j-1-q\ge n$)
$$\bbox[5px,border:2px solid #00A000]{
G_{n, j} = \frac{(\frac{1}{2} n(n+1))^{j-1} n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)} 
- [[j\gt n]] n! \sum_{q=0}^{j-1-n} \left(\frac{1}{2}n(n+1)\right)^q
{j-1-q\brace n}.}$$
</s>