I was recently reminded of the Taylor series $\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty{2n\choose n}x^n$ and got wondering whether this can be proven combinatorially.
Of course the standard proof would be to look at the derivatives of $f(x)=\frac1{\sqrt{1-4x}}=(1-4x)^{-1/2}$. \begin{align*} f'(x)&=2\cdot(1-4x)^{-3/2} \\f''(x)&=4\cdot3\cdot(1-4x)^{-5/2} \\f'''(x)&=8\cdot3\cdot5\cdot(1-4x)^{-7/2} \\f^{(n)}(x)&=2^n\cdot1\cdot3\cdot5\cdot...\cdot(2n-1)\cdot(1-4x)^{-(2n+1)/2} \\&=2^n\frac{(2n)!}{n!2^n}(1-4x)^{-(2n+1)/2} \\&=\frac{(2n)!}{n!}(1-4x)^{-(2n+1)/2} \end{align*} This gives $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty{2n\choose n}x^n.$$
This is all great, but we can also look at it a different way. Note $(f(x))^2=\frac1{1-4x}=\sum_{n=0}^\infty4^nx^n$. But now note what this is saying about the Taylor series. $$\left(\sum_{n=0}^\infty{2n\choose n}x^n\right)^2=\sum_{n=0}^\infty\left(\sum_{k=0}^n{2k\choose k}{2(n-k)\choose n-k}\right)x^n=\sum_{n=0}^\infty4^nx^n$$ This gives us the identity $$\sum_{k=0}^n{2k\choose k}{2(n-k)\choose n-k}=4^n$$ which I find very curious.
My question is if there is a combinatorial proof of the final identity. This would then also give a combinatorial proof of the original Taylor series.
The way I look at the term ${2n\choose n}$ is as the amount of paths from $(0,0)$ to $(n,n)$ by only going right by one or up by one at every step. Then a product such as ${2k\choose k}{2(n-k)\choose n-k}$ would be the amount of such paths that cross the point $(k,k)$. This is because we can simply join the paths to $(k,k)$ and $(n-k,n-k)$. But this means that every path from $(0,0)$ to $(n,n)$ gets counted in the product ${2k\choose k}{2(n-k)\choose n-k}$ if and only if it crosses $(k,k)$.
In conclusion, the sum $\sum_{k=0}^n{2k\choose k}{2(n-k)\choose n-k}$ can be combinatorially seen as the sum over all paths from $(0,0)$ to $(n,n)$ of the amount of times it touches the line $y=x$. The goal is then to show that this exactly equals $4^n$.
