TITLE: Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$
QUESTION [18 upvotes]: How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$?
I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me.
(This series arises in studying the first passage time of a simple random walk.)

REPLY [13 votes]: Or, by definition.
\begin{eqnarray*}
{-1/2\choose n}&=&{(-1/2)(-1/2-1)(-1/2-2)\cdots(-1/2-[n-1])\over n!}\cr
&=&{(-1)^n\over 2^n} {(1)(3)(5)\cdots(2n-1)\over n!}\cr
&=&{(-1)^n\over 2^n} {(1)(3)(5)\cdots(2n-1)\over n!}\cdot{2^n n!\over 2^n n!}\cr 
&=&{(-1)^n\over 4^n} {2n\choose n}.
\end{eqnarray*}