Let $n$ be a positive integer and $N=\displaystyle\sum_{k=0}^{n}(-1)^k\binom{n}{k}^2$. Show that $N=0$ if $n$ is odd, and $N=(-1)^{\frac{n}{2}}\displaystyle\binom{n}{\frac{n}{2}}$ if $n$ is even.
The solution for this problem is available for $0.99.
You can also purchase a pass for all available solutions for $99.