Proof that $\sum_{k=1}^{n}{k^{3}\binom{n}{k}^{2}}=2n\binom{n}{2}\binom{2n-3}{n-2}+n^{2}\binom{2n-2}{n-1}$ I want to prove that the following equation is correct:
$$
\sum_{k=1}^{n}{k^{3}\binom{n}{k}^{2}}=2n\binom{n}{2}\binom{2n-3}{n-2}+n^{2}\binom{2n-2}{n-1}
$$
I used combinatorial proof. There are two companies $A$ and $B$ and each have $n$ employees. Company $A$ and $B$ swap some employees but the number of employees in each companies remain the same.  Afterwards company $A$ elect a technical manager and company $B$ elect a technical manager and marketing manager, can be the same person. However, manager in $A$ used to be in $B$ and manager(s) in $B$ used to be in $A$.
Method 1
Select $k$ employees from each company to be swapped, $k\in\{1,2,…,n\}$. From $k$ ex $B$ select one technical manager and from $k$ ex $A$ select technical and marketing manager. The number of possibilities is given below
$$
\sum_{k=1}^{n}{\binom{n}{k}^{2}\cdot k\cdot k^{2}}
$$
Method 2
Select two people from $A$ to be marketing and technical manager respectively then move them to $B$. Select one people from $B$ to be technical manager and move them to $A$. Then complete the companies to make a total of $n$ employees per employees. The number of possibilities is given below
$$
2\binom{n}{2}\cdot\binom{n}{1}\cdot\binom{2n-3}{n-2}
$$
If technical and marketing manager is the same person, select one person from $A$ to be both and move him to $B$. Select one person from $B$ to be technical manager and move him to $A$. Then complete the company member. The number of possibilities is given below
$$
n\cdot n\cdot\binom{2n-2}{n-1}
$$
Since I counted the same possibilities using different methods, the two expressions must be equal.
I want to know if my solution is correct and if anyone has alternative e.g. generating function or algebra solution
 A: Your proof is fine. Here it is a direct approach. Note that
$$k^3=k(k-1)(k-2)+3k(k-1)+k.$$
Hence, by Vandermonde's identity, we have
$$\begin{align}
\sum_{k=1}^{n}k^{3}\binom{n}{k}^{2}&=\sum_{k=1}^{n}k(k-1)(k-2)\binom{n}{k}\binom{n}{n-k}+3\sum_{k=1}^{n}k(k-1)\binom{n}{k}\binom{n}{n-k}\\
&\qquad\qquad  +\sum_{k=1}^{n}k\binom{n}{k}\binom{n}{n-k}\\
&=n(n-1)(n-2)\sum_{k=3}^{n}\binom{n-3}{k-3}\binom{n}{n-k}+3n(n-1)\sum_{k=2}^{n}\binom{n-2}{k-2}\binom{n}{n-k}\\
&\qquad\qquad +n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{n-k}\\
&=n(n-1)(n-2)\binom{2n-3}{n-3}+3n(n-1)\binom{2n-2}{n-2}+n\binom{2n-1}{n-1}\\
&=2n\binom{n}{2}\binom{2n-3}{n-2}+n^{2}\binom{2n-2}{n-1}.
\end{align}$$
A: Seems good to me. For an algebraic approach consider $k\binom{n}{k}=n\binom{n-1}{k-1}$ so
$$\sum k\cdot k\cdot \binom{n}{k}\cdot k\cdot \binom{n}{k}=n^2\sum k\binom{n-1}{k-1}^2=n^2\sum (k-1)\binom{n-1}{k-1}^2+n^2\sum \binom{n-1}{k-1}^2,$$
the second sum evaluates to $n^2\binom{2n-2}{n-1}$ using Vandermonde.
For the first one use again the first identity to get
$$n^2(n-1)\sum \binom{n-2}{k-2}\binom{n-1}{k-1}=n^2(n-1)\binom{2n-3}{n-2}.$$
A: We can evaluate the following general sum:
$$\sum_{k=1}^n k^\ell {n\choose k}^2.$$
We get
$$\ell! [w^\ell] \sum_{k=0}^n \exp(kw) {n\choose k}
{n\choose n-k}
= \ell! [w^\ell] [z^n] (1+z\exp(w))^n (1+z)^n
\\ = \ell! [w^\ell] [z^n] (1+z+z(\exp(w)-1))^n (1+z)^n
\\ = \ell! [w^\ell] [z^n] \sum_{k=0}^n {n\choose k}
z^k (\exp(w)-1)^k (1+z)^{2n-k}
\\ = \sum_{k=0}^n {n\choose k} {2n-k\choose n-k}
k! {\ell \brace k}.$$
Note that when $n\gt \ell$ we may lower to $\ell$ due to the Stirling
number. On the other hand when $n\lt \ell$ we may raise to $\ell$ due to
the first binomial coefficient. We get
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=1}^\ell {2n-k\choose n} n^{\underline{k}} {\ell\brace k}.}$$
We find e.g. for $\ell = 3$
$${2n-1\choose n} n {3\brace 1}
+ {2n-2\choose n} n (n-1) {3\brace 2}
+ {2n-3\choose n} n (n-1) (n-2) {3\brace 3}.$$
