Proving an Identity involving $4^N$ I am trying to prove the following identity:
$$\sum_{k=0}^N\left({2 \, N - 2 \, k \choose N - k}{2 \, k \choose k}\right)=4^N$$
I have tried writing $4^N=2^{2N}=(1+1)^{2N}=(1+1)^N(1+1)^N$, and expanding each of these as a binomial expansion, but I have found nothing but dead ends so far. Any ideas?
I am currently working through a Ch. 3 "Generating Functions" from  Analysis of Algorithms by Sedgewick/Flajolet. This is problem #30.
Thanks. 

Let, $f(x)=\displaystyle\sum_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$. Then your LHS will be coefficient of $x^N$ of $f^2(x)=(1-4x)^{-1}$ which is $4^N$=RHS.
