<s>[INST] 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.
 [/INST] 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,
$$
 = n\binom{2n}{4} - \binom{2n}{5} + \binom{n}{5}
$$
</s>