Find $\sum_{k=1}^{2n} k {2n\choose k}^2 (-1)^{k+1}$. [duplicate]

Question

I need to find $$S = {2n\choose 1}^2 -2 {2n\choose 2}^2 + ... - 2n{2n\choose 2n}^2= \sum_{k=1}^{2n} k {2n\choose k}^2 (-1)^{k+1}$$, given $$\sum_{k=1}^{2n} k {2n\choose k} x^{k-1} = 2n(1+ x)^{2n -1}$$

Using $$-(1-x)^{2n} + 1 = \sum_{k=1}^{2n} {2n \choose k} (-1)^{k+1} x^k$$,

$$(1- (1-x)^{2n}) (2n (1+x)^{2n-1}) = \sum_{k=1}^{4n}x^k\sum_{r=0}^k {2n\choose r} (-1)^{r+1} (k - r) {2n \choose k-r}$$

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

So coefficient of $x^{2n}$ on the LHS is $-2n{2n - 1\choose n} (-1)^n = -n{2n \choose n} (-1)^n$

Coefficient of $x^{2n}$ on the RHS,

$$2n \sum_{r=0}^{2n} {2n\choose r}^2 (-1)^{r+1} - S = - 2n{2n \choose n}(-1)^n - S$$

Equating the coefficient of both sides gives $$S = -n{2n \choose n} (-1)^n$$.

I would like to know different methods for doing this and similar problems as the method I used is cumbersome and prone to miscalculation.

Answer 1

$[x^n]:f(x)$ denote the coefficient of $x^n$ in the function $f(x)$ \begin{eqnarray*} \binom{2n}{k} =\binom{2n}{2n-k}=[x^{2n-k}]: (1+x)^{2n}=[x^{2n}]:x^k(1+x)^{2n}. \end{eqnarray*} So your sum can be written as \begin{eqnarray*} \sum_{k=1}^{2n} k \binom{2n}{k}^2 (-1)^{k+1} &=& [x^{2n}]: (1+x)^{2n} \sum_{k=1}^{2n} k \binom{2n}{k} (-1)^{k}x^k \\ &=& [x^{2n}]:2n x (1-x)^{2n-1} (1+x)^{2n} \\ &=& [x^{2n}]:2n x(1+x) (1-x^{2})^{2n-1} . \end{eqnarray*} We only need to consider the even powers of $x$, so let $y=x^2$ and we have \begin{eqnarray*} \sum_{k=1}^{2n} k \binom{2n}{k}^2 (-1)^{k} &=& [y^n]: 2n y (1-y)^{2n-1} =2n \binom{2n-1}{n-1} (-1)^{n-1}. \end{eqnarray*} So your answer is correct.