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.

  • $\begingroup$ You have a sign error somewhere, the answer should be $(-1)^n n{2n\choose n}$. $\endgroup$ May 1, 2019 at 21:00
  • $\begingroup$ @Jean-ClaudeArbaut For $n=2$ I got the sum as $-12 = - 2 {4 \choose 2} (-1)^{2}$. It should be $+12$ if I had a sign error. $\endgroup$ May 1, 2019 at 21:55
  • $\begingroup$ Ah, understood. I got the formula from the title, but it's not the same as in the question. So there was indeed a sign error, in the title. $\endgroup$ May 1, 2019 at 22:07
  • $\begingroup$ Your question is a special case of the one I linked. In the linked question, set $k\gets 2n$ and $n\gets 2n$ to recover (the negation of) your sum. $\endgroup$ May 1, 2019 at 23:21

1 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.


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