Binomial coefficients identity Prove algebraically or otherwise:
$$\sum \limits_{r=0}^n {2r \choose r} {2n-2r \choose n-r} = 4^n $$
where ${n \choose r}$ denotes the usual binomial coefficient. I think there is a combinatorial proof using path counting arguments, but I haven't been able to find it.

Start with
$$(1+kx)^\alpha=1+\sum\limits_{j=1}^\infty{\alpha(\alpha-1)\cdots(\alpha-j+1)\over j!}k^jx^j$$
With $\alpha=-{1\over 2}$ and $k=-4$ the general term of the above series is
$${2k\choose k}={(k+1)\cdots 2k\over k!}$$
$\sum \limits_{r=0}^n {2r \choose r} {2n-2r \choose n-r}$ is just the coefficient of $x^n$ in the series Cauchy product
$$\left(\sum\limits_{k=0}^\infty{2k\choose k}x^k\right)^2$$
Now keep in mind that 
$$\sum_{k=0}^\infty{2k\choose k}x^k={1\over\sqrt{1-4x}}$$
So our sum is the coefficient of $x^n$ in the series expansion of
$${1\over 1-4x}=\sum\limits_{k=0}^\infty 4^kx^k$$
so it is $4^n$ as requested.
