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

$[x^n]:f(x)$ 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.
