[Math] Alternating sum of squares of binomial coefficients

Question

I know that the sum of squares of binomial coefficients is just ${2n}\choose{n}$ but what is the closed expression for the sum ${n\choose 0}^2 – {n\choose 1}^2 + {n\choose 2}^2 + \cdots + (-1)^n {n\choose n}^2$?

Answer 1

$$(1+x)^n(1-x)^n=\left( \sum_{i=0}^n {n \choose i}x^i \right)\left( \sum_{i=0}^n {n \choose i}(-x)^i \right)$$

The coefficient of $x^n$ is $\sum_{k=0}^n {n \choose n-k}(-1)^k {n \choose k}$ which is exactly your sum.

On another hand:

$$(1+x)^n(1-x)^n=(1-x^2)^n=\left( \sum_{i=0}^n {n \choose i}(-1)^ix^{2i} \right)$$

Thus, the coefficient of $x^n$ is $0$ if $n$ is odd or $(-1)^{\frac{n}2}{n \choose n/2}$ if $n$ is even.