The summation $\sum_{k=0}^{2n} (-1)^k \frac{{4n \choose 2k}}{{2n \choose k}}=\frac{1}{1-2n}$

By using W-Z method, we show a more general identity: for $n\geq 2$, $$\sum_{k=0}^{n} (-1)^k \frac{\binom{2n}{2k}}{\binom{n}{k}}=\frac{1+(-1)^n}{2(1-n)}.$$

Let $$F(n,k)= (-1)^k \frac{\binom{2n}{2k}}{\binom{n}{k}}(n-1).$$ Then for $k=0,1,\dots,n$, $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$ with $$G(n,k)=(-1)^k\frac{\binom{2n}{2k-2}}{\binom{n+1}{k}}(n+1).$$ Hence $$\begin{align} \sum_{k=0}^{n+1}F(n+1,k)-\sum_{k=0}^{n}F(n,k)&=F(n+1,n+1)+ \sum_{k=0}^{n}(G(n,k+1)-G(n,k))\\ &=F(n+1,n+1)+G(n,n+1)-G(n,0)\\ &=(-1)^{n+1}n+(-1)^n(n+1)-0\\ &=(-1)^n. \end{align}.$$ Finally, for $n\geq 2$, $$\sum_{k=0}^{n} (-1)^k \frac{\binom{2n}{2k}}{\binom{n}{k}}=\frac{1}{n-1}\sum_{k=0}^{n}F(n,k)=\frac{1}{n-1}\sum_{k=2}^{n}(-1)^{k-1}=\frac{1+(-1)^n}{2(1-n)}.$$