Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$

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.)

Solution 1:

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*}

Solution 2:

The key identities are the duplication formula for the factorial (which I'll recast in a more convenient format):

$$\binom{2n}{n}=\frac{4^n}{\sqrt \pi}\frac{\left(n-\frac12\right)!}{n!}$$

and the reflection formula

$$\left(-n-\frac12\right)!\left(n-\frac12\right)!=(-1)^n\pi$$

Making the appropriate replacements, we obtain

$$\binom{2n}{n}=(-4)^n\frac{\sqrt \pi}{n!\left(-n-\frac12\right)!}=(-4)^n\frac{\left(-\frac12\right)!}{n!\left(-n-\frac12\right)!}=(-4)^n\binom{-\frac12}{n}$$

You can proceed from that...