Prove this binomial identity

Question

Prove the binomial identity $$\sum\limits_{i=0}^{n}(-1)^i {n\choose{i}}{2i\choose{n}}=(-2)^n$$

Answer 1

Using the notation $[x^n]: f(x)$ to represent the coefficient of $x^n$ in the function $f(x)$. So for example \begin{eqnarray*} \binom{2i}{n} = [x^n]: (1+x)^{2i}. \end{eqnarray*}

\begin{eqnarray*} \sum_{i=0}^{n} (-1)^i \binom{n}{i} \binom{2i}{n} &=& [x^n] : \sum_{i=0}^{n} (-1)^i \binom{n}{i} (1+x)^{2i} \\ &=& [x^n] : (1- (1+x)^{2} )^n \\ &=& [x^n] :(-1)^n (2x +x^2 )^n\\ &=&(-2)^n. \end{eqnarray*}