Evaluating sum of a combination series Please help me evaluate the following series.

$$\sum_{k=0}^{n} k{n\choose k}^2.$$

 A: Method 1 (Generating function oriented)
$$
\begin{align}
\sum_{k=0}^n k \binom{n}{k}^2 
\stackrel{\color{blue}{^{[1]}}}{=} 
& \sum_{k=0}^n\sum_{l=0}^n k\binom{n}{k}\binom{n}{l}\delta_{kl}\\
\stackrel{\color{blue}{^{[2]}}}{=} 
& \sum_{k=0}^n\sum_{l=0}^n k\binom{n}{k}\binom{n}{l}\left(\int_0^{2\pi}e^{i(k-l)\theta}\frac{d\theta}{2\pi}\right)\\
\stackrel{\color{blue}{^{[3]}}}{=} 
& \lim_{s\to 1} s\frac{d}{ds}\left[\sum_{k=0}^n\sum_{l=0}^n \binom{n}{k}\binom{n}{l}\left(s^k\int_0^{2\pi}e^{i(k-l)\theta}\frac{d\theta}{2\pi}\right)\right]\\
= & \lim_{s\to 1} s\frac{d}{ds}\left[\int_0^{2\pi} ( 1 + s e^{i\theta})^n( 1 + e^{-i\theta})^n\frac{d\theta}{2\pi}\right]\\
= & n \int_0^{2\pi} e^{i\theta}( 1 + e^{i\theta})^{n-1}( 1 + e^{-i\theta})^n\frac{d\theta}{2\pi}\\
= & n \int_0^{2\pi} e^{-i(n-1)\theta} ( 1 + e^{i\theta} )^{2n-1} \frac{d\theta}{2\pi}\\
\stackrel{\color{blue}{^{[4]}}}{=} 
& n \binom{2n-1}{n-1}
\end{align}$$
Notes


*

*$\color{blue}{[1]}$ $\delta_{kl}$ is the Kronecker delta.

*$\color{blue}{[2]}$ Given two sequences $a_k$, $b_k$ and corresponding generating functions
$$a(z) = \sum_{k} a_k z^k\quad\text{ and }\quad b(z) = \sum_{k} b_k z^k.$$
the introduction of a Kronecker delta followed by a integral representation:
$$\delta_{kl} = \int_0^{2\pi} e^{i(k-l)\theta} \frac{d\theta}{2\pi}$$
is one trick to obtain a generating function for the product sequence $a_kb_k$:
$$\sum_{k}a_kb_k z^k = \int_{0}^{2\pi} a(\sqrt{z}e^{i\theta}) b(\sqrt{z} e^{-i\theta}) \frac{d\theta}{2\pi}$$

*$\color{blue}{[3]}$ The introduction of an auxiliary variable, $s$ in this case,
and then replace $k$ by $s\frac{d}{ds} s^k$ is another common trick to deal with
combinatorial sums that involve binomial coefficients.

*$\color{blue}{[4]}$ Let $z = e^{i\theta}$. The $e^{-i(n-1)\theta}$ factor in the integrand pick up and only pick up the coefficient of $z^n$ in the polynomial $(1+z)^{2n-1}$. We then apply binomial theorem to $(1+z)^{2n-1}$ to obtain the last expression.


Method 2 (Combinatorial)
Notice $\displaystyle\quad\binom{n}{k} = \binom{n}{n-k}\quad$, we have:
$$\sum_{k=0}^n k \binom{n}{k}^2 = \sum_{k=0}^n (n-k)\binom{n}{n-k}^2 
= \sum_{k=0}^n(n-k)\binom{n}{k}^2$$
This implies 
$$\sum_{k=0}^n k \binom{n}{k}^2 = \frac{n}{2}\sum_{k=0}^n \binom{n}{k}^2
= \frac{n}{2}\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k}$$
The last sum in R.H.S has a combinatorial realization. 
If you are given $2n$ balls and asked to pick $n$ balls out of them. One way to count the number of possibilities is first split the $2n$ balls into two groups of $n$ balls. You then pick $0 \le k \le n$ from the first group and then $n-k$ balls from the second group.
For each $k$, the number of possibilities is $\displaystyle \binom{n}{k} \binom{n}{n-k}$.
Since we know the sum of these numbers will give us the total number of possibilities $\displaystyle \binom{2n}{n}$, we obtain following identity:
$$\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} = \binom{2n}{n}$$
Using this, we can conclude
$$\sum_{k=0}^n k \binom{n}{k}^2 = \frac{n}{2}\binom{2n}{n} = n\binom{2n-1}{n-1}$$
A: I take it you mean the sum is over $k$. 
You can start by writing $k\binom{n}{k} = n\binom{n-1}{n-k}$. This effectively reduces the problem to evaluating $\sum_k \binom{n-1}{n-k}\binom{n}{k}$. Then you can apply a Vandermonde convolution identity. 
