Closed form of a sum

Let $n$ be a non negative integer. Find a closed expression for:

$$\sum_{k=0}^{n}(-1)^k 2^k \binom{n}{k}\binom{2n-k}{n}$$

(Michael Bataille, Rouen, France)

Solution  ( José H. Nieto, Universidad del Zulia, Maracaibo, Venezuela)

Let $\langle x^k \rangle f(x)$ denote the coefficient of $x^k$ in the polynomial of $f(x)$. Clearly , $\langle x^{n-k} \rangle f(x) = \langle x^n \rangle x^k f(x)$ and $\sum_{k} \langle x^k \rangle f(x) y^k = f(y)$. Now we have that:

$$\langle x^k \rangle \left ( 1-2x \right )^n = (-2)^k \binom{n}{k}= (-1)^k 2^k \binom{n}{k}$$

and 

$$\binom{2n-k}{n}= \binom{2n-k}{n-k}= \langle y^{n-k} \rangle \left ( 1+y \right )^{2n-k}=\langle y^n \rangle y^k \left ( 1+y \right )^{2n-k}$$

Hence:

$$\begin{aligned}
\sum_{k=0}^{n}(-1)^k 2^k \binom{n}{k}\binom{2n-k}{n} &=\sum_{k=0}^{n}\langle x^k \rangle \left ( 1-2x \right )^n \langle y^n \rangle y^k \left ( 1+y \right )^{2n-k} \\
 &= \langle y^n \rangle \left ( 1+y \right )^{2n}\sum_{k=0}^{n}\langle x^k \rangle \left ( 1-2x \right )^n \left ( \frac{y}{1+y} \right )^k\\
 &= \langle y^n \rangle \left ( 1+y \right )^{2n} \left ( 1 - \frac{2y}{1+y} \right )^n \\
 &= \langle y^n \rangle \left ( 1+y \right )^n \left ( 1-y \right )^n\\
 &= \langle y^n \rangle \left ( 1-y^2 \right )^n\\
 &= \left\{\begin{matrix}
 0&, &{\rm if} \; n  \; {\rm is \; odd} \\
 (-1)^{n/2}\binom{n}{n/2}&, & {\rm if} \; n \; {\rm is even}
\end{matrix}\right.
\end{aligned}$$

Source: Crux Mathematicorum