It's not difficult to show that

$$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$

On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$.  Squaring the first power series and comparing terms gives us

$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$

that is,

$$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$

My question: is there a more direct, combinatorial proof of this identity?  I've been racking my brains trying to come up with one but I'm not having much success.

It is possible to give a direct combinatorial proof, but it is quite difficult to find it.

One possibility is to use paths between points with integer coordinates and steps $(1,1)$ and $(1,-1)$.

1) $\binom{2n}{n}$ counts all paths from $(0,0)$ to $(2n,0)$.

2) $2^{2n}$ counts all paths starting from $(0,0)$ with $2n$ steps.

3) $\binom{2n}{n}$ counts all paths with $2n$ steps that never touch the  $x$-axis again after the start. (This one is not obvious, but can be proved 
with a bijection.)

Now you can conclude that all paths are a concatenation of a path that returns a certain number of times to the $x$-axis and a path that never does.

Note that the main difficulty here was that the two binomial coefficients are interpreted differently.

Edited to add reference: 
In Richard P. Stanley: Enumerative Combinatorics Volume 1, Chapter 1, Solution to exercice 2c the following reference is given:


  The problem of giving a combinatorial proof was raised by P. Veress and solved by G. Hajos in the 1930s. A recent proof appears in D.J. Kleitman, Studies in Applied Math. 54 (1975), 289 - 292. See also M. Sved, Math. Intelligencer, vol.6, no. 4 (1984), 44-45.


But I have not looked to check which article gives the proof I have outlined above.
