Write out what $\binom{-1/2}{n}$ means; i.e., $ \begin{align} \binom{-1/2}{n} &= \frac{(-1/2)(-3/2) \cdots ((-2n+1)/2)}{n!} = \frac{(-1)^n}{2^n} \frac{(1)(3) \cdots (2n-1)}{n!} \\ &= \frac{(-1)^n}{2^n} \frac{(2n)!}{2(4) \cdots (2n)n!} = \left(\frac{-1}{4}\right)^n \frac{(2n)!}{n!n!} \\ &= \left(\frac{-1}{4}\right)^n \binom{2n}{n}. \end{align} $ This should be enough for you to be able to find $\sum_{n=0}^{\infty} \binom{2n}{n} a^n$.
Are you sure you mean
$\sqrt{1+x}$, though? That would give
$\sqrt{1+x} = \sum_{n=0}^{\infty} \binom{1/2}{n} x^n$. Then, following through the same argument as above you would obtain
$\binom{1/2}{n} = \frac{-1}{2n-1} \left(\frac{-1}{4}\right)^n \binom{2n}{n}$, which would be a bit more difficult to deal with because of the
$2n-1$ in the denominator.