<s>[INST] Sum of square of binomial coeffcient with positive and negative terms Finding $\displaystyle \binom{2n}{1}^2-2\binom{2n}{2}^2+3\binom{2n}{3}^2-\cdots \cdots -2n\binom{2n}{2n}^2.$
What I've tried:
$$(1-x)^{2n}=\binom{2n}{0}-\binom{2n}{1}x+\binom{2n}{2}x^2+\cdots \cdots +\binom{2n}{2n}x^{2n}$$
$$-2n(1-x)^{2n-1}=-\binom{2n}{1}+2\binom{2n}{2}x-3\binom{2n}{3}x^2+\cdots +n\binom{2n}{2n}x^{2n-1}$$

Sum notation:
$$\sum_{k=0}^{2n} (-1)^{k-1}k\binom{2n}{k}^2$$

 [/INST] Starting from
$$\sum_{k=0}^{2n} (-1)^{k-1} k {2n\choose k}^2
= \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
k {2n\choose k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose k-1}
= 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
{2n-1\choose 2n-k}
\\ = 2n \sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k}
[z^{2n-k}] (1+z)^{2n-1}
\\ = 2n [z^{2n}] (1+z)^{2n-1}
\sum_{k=1}^{2n} (-1)^{k-1} {2n\choose k} z^k
\\ = 2n [z^{2n}] (1+z)^{2n-1}
\left(1+\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k\right).$$
Now $2n [z^{2n}] (1+z)^{2n-1}$ is zero, so we may continue with
$$2n [z^{2n}] (1+z)^{2n-1}
\sum_{k=0}^{2n} (-1)^{k-1} {2n\choose k} z^k
\\ = - 2n [z^{2n}] (1+z)^{2n-1} (1-z)^{2n}
= - 2n [z^{2n}] (1-z^2)^{2n-1} (1-z)
\\ = - 2n [z^{2n}] (1-z^2)^{2n-1}
= - 2n [z^{n}] (1-z)^{2n-1}.$$
This is
$$(-1)^{n+1} \times 2n \times {2n-1\choose n}
= (-1)^{n+1} \times 2n \times {2n\choose n}
\frac{n}{2n}$$
for an answer of
$$\bbox[5px,border:2px solid #00A000]{
(-1)^{n+1} \times n \times {2n\choose n}.}$$
</s>