Proving $\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $

So the left hand side has been changed to give the equation $$ \sum_{k=0}^nk\binom nk^2 = n\binom{2n-1}{n-1}. $$ This becomes slightly easier if you replace one factor $\binom nk$ by the equivalent $\binom n{n-k}$, to give as equation to prove $$ \sum_{k=0}^nk\binom nk\binom n{n-k} = n\binom{2n-1}{n-1}. $$ Suppose you have a group consisting of $n$ boys and $n$ girls. A team of $n$ has to be formed out of them, and a captian designated of the team, which has to be a girl (I did not succeed in thinking of a less discriminatory example). On one hand, one can fix the number $k$ of girls and $n-k$ of boys on the team first, then choose the team in one of $\binom nk\binom n{n-k}$ ways and a captain in one of $k$ ways, so the left hand side counts the number of possible selections. On the other hand one could start out to choose one of the $n$ girls as captain, and let her choose $n-1$ team mates among the remaining $2n-1$ children; this is counted by the right hand side.

Here's a relatively simple way to do this with just algebra.

We want to sum

$$S=\sum_{k=0}^n k\binom{n}{k}^2 = \sum_{k=0}^n k\binom{n}{k}\binom{n}{n-k} .$$

Note, using symmetry, that this is equal to

$$\sum_{k=0}^n (n-k)\binom{n}{k}\binom{n}{n-k}.$$

Adding these, we get

$$2S=n\sum_{k=0}^n \binom{n}{k}\binom{n}{n-k}.$$

This means it suffices to evaluate

$$\sum_{k=0}^n \binom{n}{k}\binom{n}{n-k}.$$ This sum is $\binom{2n}{n}$. For justification, I quote another answer:

$$(1+x)^n(x+1)^n=(1+x)^{2n}$$

$$\left(\sum_{0\le r\le n}\binom nr x^r \right)\left(\sum_{0\le r\le > n}\binom nr x^{n-r}\right)=\sum_{0\le r\le 2n}\binom {2n}rx^r$$

Compare the coefficients of $x^n.$

But wait? We have $2n$ and $n$ where we want $2n-1$ and $n-1$! This is not a problem. Recall the "in-n-out" formula:

$$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}.$$

It’s false as stated. Take $n=3$:

$$\sum_{k=0}^3k\binom3k=0\cdot1+1\cdot3+2\cdot3+3\cdot1=12\ne 30=3\binom52$$

In fact

$$\sum_{k=0}^nk\binom{n}k=\sum_{k=0}^nn\binom{n-1}{k-1}=n\sum_{k=1}^n\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n-1}k=n2^{n-1}\;.$$

A combinatorial proof of this corrected identity isn’t too hard to find. Start with a pool of $n$ people. For each $k\in\{0,\dots,n\}$, the term $k\binom{n}k$ is the number of ways to choose a $k$-person team and then pick one member of the team to be captain. Summing over $k$ gives the total number of ways to pick a team of any size up through $n$ and then choose one of its members to be captain. Alternatively, we can first pick any of the $n$ people to be the team captain, and we can then pick any subset of the remaining $n-1$ people to make up the rest of the team; this combination of selections can be made in $n2^{n-1}$ ways.

Added: For the corrected question, imagine that you have $n$ women and $n-1$ men. There are $\binom{2n-1}n=\binom{2n-1}{n-1}$ ways to choose a team of $n$ people from that group and $n$ ways to choose one of them to be captain, so we can choose a team of $n$ with its captain in $n\binom{2n-1}{n-1}$ ways. Alternatively, we can choose $k-1$ men in $\binom{n-1}{k-1}$ ways and $n-k$ women in $\binom{n}{n-k}=\binom{n}k$ ways to get our team of $n$, and we can then choose a captain in $n$ ways, for a total of

$$n\binom{n-1}{k-1}\binom{n}k=k\binom{n}k^2$$

teams with captain having exactly $k-1$ men. Here I'm using the identity $k\binom{n}k=n\binom{n-1}{k-1}$, which is easily verified algebraically or combinatorially. Now just sum over the possible values of $k$ to get the result.