<s>[INST] How to prove that $\sum_{k=0}^n{(-1)^k{4n-2k\choose 2n}{2n\choose k}}=2^{2n}$? $$\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}=2^{2n}$$
I know the correctness of this formula, but how can I prove it?
Thanks for your help.
 [/INST] 
Prove that$$\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}=2^{2n}$$

Consider the following binomial expansion
\begin{align*}
(x^2-1)^{2n}&=\sum_{k=0}^{2n}(-1)^kx^{4n-2k}{2n\choose k}
\end{align*}
Put $x=1$ after differentiating both sides $2n$ times with respect to $x$ and you are done!
(Stop reading and try it yourself for a "confidence-booster")
\begin{align*}
2x(2n)(x^2-1)^{2n-1}&=\sum_k(-1)^k(4n-2k)x^{4n-2k-1}{2n\choose k}\\
2(2n)(x^2-1)^{2n-1}+4x^2(2n)(2n-1)(x^2-1)^{2n-2}&=\sum_{k}(-1)^k(4n-2k)x^{4n-2k-1}{2n\choose k}\\
\end{align*}
Let's neglect terms like the leftmost one in the above expression, for we are eventually going to put $x=1$ and since $2n-k>2n-k-1$. So,
\begin{align*}
\cdots+2^2x^2(2n)(2n-1)(x^2-1)^{2n-2}&=\sum_k(-1)^k(4n-2k)x^{4n-2k-1}{2n\choose k}\\
\cdots+2^3x^3(2n)(2n-1)(2n-2)(x^2-1)^{2n-3}&=\sum_k(-1)^k(4n-2k)(4n-2k-1)x^{4n-2k-2}{2n\choose k}\\
\end{align*}
and so on so that at last
\begin{align*}
2^{2n}x^{2n}(2n)!(x^2-1)^0&=\sum_k(-1)^k(4n-2k)(4n-2k-1)\cdots(2n-2k+1)x^{2n-2k}{2n\choose k}\\
\end{align*}
Now, what is the greatest value $k$ can take so that the latter terms aren't zero due to differentiation $\rightarrow2n-2k=0\Rightarrow k=n$.
So, after putting $x=1$, we get
\begin{align*}
2^{2n}(2n)!&=\sum_{k=0}^{n}(-1)^k(4n-2k)(4n-2k-1)\cdots(2n-2k+1){2n\choose k}\\
2^{2n}&=\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}
\end{align*}
</s>