Prove alternating sum of binomial coefficients squared by induction

Question

I am trying to find the sum

$${100\choose 0}^2-{100\choose 1}^2+{100\choose 2}^2- \dots+ {100\choose 100}^2$$

After the first few examples for small numbers this appears to be

$$\sum_{k=0}^{n}{(-1)^k {n\choose k}^2}=(-1)^\frac{n}{2}{n\choose \frac{n}{2}}$$ for all even $n$s. (For odd $n$ this is obviously $0$ because of the symmetry ${n\choose k}={n\choose{n-k}}$).

I tried to prove it by induction, but it is too messy and I am not sure that's the right direction. What I did is write it in terms of $m=\frac{n}{2}$ so then the induction step will be just $+1$, although that didn't help very much.

Thanks in advance for any help.

Answer 1

Consider two binomial series $(1+x)^{n}$ and $(1-x)^{n}$. Note that the terms in the sum are the product of the coefficients of the terms in these expansions.

$$\begin{aligned}\sum_{k=0}^{n}(-1)^{k}\left[{n\choose k}\right]^{2}&=\text{constant term in }(1+x)^{n}\left(1-\frac{1}{x}\right)^{n}\\&=\text{coeff. of } x^{n} \text{ in }(x^2-1)^{n}=(-1)^{\frac{n}{2}}{n\choose \frac{n}{2}}\ \square\end{aligned}$$