How to prove this indentity $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$

Question

I don't know how to prove this identity: $\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-\binom{100}{3}^2+...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}$