Show that $$\displaystyle\sum_{k=0}^{2n}(-1)^k \dbinom{2n}{k}^2 = (-1)^n\dbinom{2n}{n}$$ where $n \in \mathbb{N}$
Induction didn't work, and the formula $\displaystyle\sum_{k=0}^{2n}\dbinom{2n}{k}^2 = \dbinom{4n}{2n}$ wasn't very usefull. $(1+x)^{2n} = \displaystyle\sum_{r=0}^{2n} \dbinom{2n}{r} x^{r}$ and $(1-x)^{2n} = \displaystyle\sum_{l=0}^{2n} \dbinom{2n}{l} (-1)^lx^{l}$. Multiplying the two gives me,
$$ (1+x)^{2n}(1-x)^{2n} = (1-x^2)^{2n} = \displaystyle\sum_{k=0}^{2n} \dbinom{2n}{k}(-1)^kx^{2k} = \left( \displaystyle\sum_{r=0}^{2n} \dbinom{2n}{r} x^{r}\right)\left( \displaystyle\sum_{l=0}^{2n} \dbinom{2n}{l} (-1)^lx^{l}\right) $$
Coefficient of $x^{2n}$ on left side will be $\dbinom{2n}{n}(-1)^n$. But how do I exactly calculate the coefficient of $x^{2n}$ in the R.H.S? Please give a detailed solution