How could I obtain
$$\sum_{k=0}^{\infty} {4k \choose 2k} \frac{z^{2k}}{2^{4k}}$$
from
$$\sum_{k=0}^{\infty} {2k \choose k} \frac{z^k}{2^{2k}}$$
which is $(1-z)^{-\frac{1}{2}}$. I can't manage to take out only the even terms of the sum... Can you help me please?
Any time you have a series $f(z)=\sum_{k=0}^\infty a_kz^k$, you get $$\sum_{k=0}^\infty a_{2k}z^{2k} = \frac{f(z)+f(-z)}{2}$$ and similarly:
$$\sum_{k=0}^\infty a_{2k+1}z^{2k+1} = \frac{f(z)-f(-z)}{2}$$