A Curious binomial identity While playing around with random binomial coefficients , I observed that the following identity seems to hold for all positive integers $n$:
$$ \sum_{k=0}^{2n} (-1)^k \binom{4n}{2k}\cdot\binom{2n}{k}^{-1}=\frac{1}{1-2n}.$$
However, I am unable to furnish a proof for it ( though this result is just a conjecture ). Any ideas/suggestions/solutions are welcome.
 A: When going through Jack's nice answer I did some intermediate steps to better see what's going on. Here is a somewhat more elaborated version, which might also be convenient for other readers.

We  obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blue}{(-1)^k\binom{4n}{2k}\binom{2n}{k}^{-1}}\\
&=\sum_{k=0}^{2n}\binom{4n}{2k}(2n+1)\int_0^1(1-x)^kx^{2n-k}\,dx\tag{1}\\
&=(2n+1)\int_0^1x^{2n}\sum_{k=0}^{2n}\binom{4n}{2k}\left(-\frac{1-x}{x}\right)^k\,dx\tag{2}\\
&=(2n+1)\int_0^1x^{2n}\cdot\frac{1}{2}
\left(\left(1+i\sqrt{\frac{1-x}{x}}\right)^{4n}+\left(1-i\sqrt{\frac{1-x}{x}}\right)^{4n}\right)\,dx\tag{3}\\
&=\frac{2n+1}{2}\int_0^1\left(\sqrt{x}+i\sqrt{1-x}\right)^{4n}+\left(\sqrt{x}-i\sqrt{1-x}\right)^{4n}\,dx\tag{4}\\
&=(2n+1)\int_{0}^{\frac{\pi}{2}}
\left[(\cos \theta+i\sin \theta)^{4n}+(\cos\theta-i\sin \theta)^{4n}\right]\cos \theta\sin \theta\,d\theta\tag{5}\\
&=(2n+1)\int_{0}^{\frac{\pi}{2}}
\left[e^{4ni\theta}+e^{-4ni\theta}\right]\cos \theta\sin \theta\,d\theta\tag{6}\\
&=(2n+1)\int_{0}^{\frac{\pi}{2}}\cos(4n\theta)\sin(2\theta)\,d\theta\tag{7}\\
&=\frac{2n+1}{2}\int_{0}^{\frac{\pi}{2}}\left[\sin( (4n+2)\theta)-\sin ((4n-2)\theta))\right]\,d\theta\tag{8}\\
&=\frac{2n+1}{2}\left[-\frac{1}{4n+2}\cos((4n+2)\theta)
+\frac{1}{4n-2}\cos((4n-2)\theta\right]_0^{\frac{\pi}{2}}\\
&=\frac{2n+1}{2}\left(\frac{1}{2n+1}-\frac{1}{2n-1}\right)\\
&\color{blue}{=\frac{1}{1-2n}}
\end{align*}

Comment:


*

*In (1) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}

*In (2) we do some rearrangements in order to apply the binomial theorem.

*In (3) we consider the even function derived from  $(1+z)^{2n}$
\begin{align*}
\sum_{k=0}^{n}\binom{2n}{2k}z^{2k}=\frac{1}{2}\left((1+z)^{2n}+(1-z)^{2n}\right)
\end{align*}
Replacing  $n$ with $2n$  and $z$ with $i\sqrt{\frac{1-x}{x}}$ and the application of de Moivre's theorem in (6) becomes plausible.

*In (4) we do some simplifications.

*In (5) we substitute $x=\cos ^2\theta, dx=-2\cos\theta\sin\theta\,d\theta$.

*In (6) we apply De Moivre's theorem  and in (7) and (8) trigonometric sum formulas.
A: $$\begin{eqnarray*} S(n) &=& (2n+1)\sum_{k=0}^{2n}(-1)^k\binom{4n}{2k}\int_{0}^{1}(1-x)^k x^{2n-k}\,dx\tag{Euler Beta}\\&=& (2n+1)\sum_{k=0}^{2n}(-1)^k\binom{4n}{2k}\int_{0}^{1}2z(1-z^2)^k z^{4n-2k}\,dz\tag{$x\mapsto z^2$}\\&=&(2n+1)\int_{0}^{\pi/2}\sin(\theta)\cos(\theta)\left[e^{4ni\theta}+e^{-4ni\theta}\right]\,d\theta\tag{$z\to\cos\theta$}\\&=&(2n+1)\int_{0}^{\pi/2}\sin(2\theta)\cos(4n\theta)\,d\theta\tag{De Moivre} \\ &=&-\frac{2n+1}{4n^2-1} = \color{blue}{-\frac{1}{2n-1}}.\tag{Simplify}
 \end{eqnarray*}$$
