Evaluating the sum $\sum\limits_k \ k\binom{n}{k}^2$ using generating functions I have to evaluate this expression $\sum\limits_k \ k\binom{n}{k}^2$ using generating function. Could you help me please? Also with some hints.
 A: I suggest that you interpret the sum as the convolution of the generating function with coefficents $k\binom{n}{k}$ and $\binom{n}{k}$.
A: Suppose we are trying to evaluate
$$\sum_{k=0}^n k {n\choose k}^2.$$
Note that
$$k{n\choose k} = \frac{n!}{(k-1)! (n-k)!}
= n \frac{(n-1)!}{(k-1)! (n-k)!} = n{n-1 \choose k-1}.$$
This means the sum is in fact
$$n\sum_{k=1}^n {n\choose k} {n-1\choose k-1}.$$
Introduce the integral representation
$${n-1\choose k-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^k} \; dz.$$
Observe that the  the integrand is entire when $k=0$  so we may extend
the sum at its lower limit to include zero, getting
$$\frac{n}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1}
\sum_{k=0}^n {n\choose k} \frac{1}{z^k} \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1}
\left(1+\frac{1}{z}\right)^n \; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^n} \; dz.$$
This can be evaluated by inspection and produces the result
$$n \times [z^{n-1}] (1+z)^{2n-1}
= n\times {2n-1\choose n-1} = n\times {2n-1\choose n}.$$
