Using the Taylor expansion for ${(1+x)}^{-1/2}$, evaluate $\sum_{n=0}^\infty \binom{2n}{n} a^n$ Using the Taylor expansion for $${(1+x)}^{-1/2}$$ we have $${(1+x)}^{-1/2}= \sum_{n=0}^\infty \binom{-1/2}{n} (x^n)$$
for $|x|<1$. 
But if $|a| <1$, how can we use the above fact to find
$$\sum_{n=0}^\infty \binom{2n}{n} a^n?$$
Thanks! Help much appreciated.

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.
