Combinatorial summation problem: Find $ \sum_{k=1}^{n} k {n+k-1 \choose 3} $ Can we get a neat form for this summation:

$$ \sum_{k=1}^{n} k {n+k-1 \choose 3}? $$

General combinatorial tricks didn't work.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\sum_{k = 1}^{n}k{n + k - 1 \choose 3} & =
\sum_{k = 1}^{n}k\bracks{z^{3}}\pars{1 + z}^{n + k - 1} =
\bracks{z^{3}}\pars{1 + z}^{n}\sum_{k = 1}^{n}k\,\pars{1 + z}^{k - 1}
\\[5mm] & =
\bracks{z^{3}}\pars{1 + z}^{n}\,
{1 - \pars{1 + z}^{n} + nz\pars{1 + z}^{n} \over z^{2}}
\\[5mm] & =
\bracks{z^{5}}\pars{1 + z}^{n} - \bracks{z^{5}}\pars{1 + z}^{2n} +
n\bracks{z^{4}}\pars{1 + z}^{2n}
\\[5mm] &=
\bbx{{n \choose 5} - {2n \choose 5} + n{2n \choose 4}} \\ &
\end{align}
A: Let's try:
$$\begin{aligned}
\sum_{k=1}^n k\binom{n+k-1}{3} &= \sum_{k=1}^n \sum_{i=1}^k \binom{n+k-1}{3} \\
&=  \sum_{i=1}^n \sum_{k=i}^n \binom{n+k-1}{3}.
\end{aligned}$$
Using the known relation
$$
\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} 
$$
you can obtain
$$\begin{aligned}
\binom{n+i-1}{4} + \sum_{k=i}^n \binom{n+k-1}{3} 
&= \binom{n+i-1}{4} + \binom{n+i-1}{3} + \binom{n+(i+1)-1}{3}+\dots + \binom{n+n-1}{3} \\
&= \binom{n+(i+1)-1}{4} +  \binom{n+(i+1)-1}{3}+\binom{n+(i+2)-1}{3}+\dots + \binom{n+n-1}{3} \\
&=\binom{n+(i+2)-1}{4} +  \binom{n+(i+2)-1}{3}+\binom{n+(i+3)-1}{3}+\dots + \binom{n+n-1}{3} \\
&= \dots \\
&=\binom{n+n-1}{4}+\binom{n+n-1}{3} \\
&= \binom{2n}{4} \\
\end{aligned}$$
so
$$
\sum_{k=1}^n k\binom{n-k+1}{3} = \sum_{i=1}^n \binom{2n}{4}-\binom{n+i-1}{4} $$
and applying the same rule as before,
$$
 = n\binom{2n}{4} - \binom{2n}{5} + \binom{n}{5}
$$
A: The terms of the sum
$$ \sum_{k=1}^{n}k\binom{n+k-1}{3} $$
are polynomials in the $k$ variable having degree $4$. Since every polynomial can be decomposed in the binomial base (for instance $k^2=2\binom{k}{2}+\binom{k}{1}$,  $k^3=6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}$ and so on), by the hockey stick identity your sum is a polynomial in the $n$ variable with degree $5$. By computing the values of the sum at $n=1,2,3,4,5,6$ you may find the coefficients of such polynomial through Lagrange interpolation. This is a bit tedious to do by hand, but it is a $<1ms$ job for a CAS:
$$ \sum_{k=1}^{n}k\binom{n+k-1}{3}=\frac{1}{120} n (1+n) \left(-24+114 n-139 n^2+49 n^3\right).$$
