# How to derive the close form of a power series with a ${2n\choose n}$ binomial coefficient? [duplicate]

In a step in a proof that the probability to return to origin in a symmetric random walk is $1$ the following combinatorics result seems important:

$$\displaystyle \sum_{n=0}^\infty{2n\choose n}\,x^n = \frac{1}{\sqrt{1-4x}}$$

I understand that the proof involves the generalized binomial theorem, but I don't know how to get about solving the interplay between the $x^n$ in the power series and the binomial coefficients.

## marked as duplicate by Brian M. Scott combinatorics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 5 '16 at 9:02

Do you know what a taylor series is? Calculate the Taylor expansion of $$f(x)=(1-4x)^{-1/2},$$

evaluated at $x_0=0$. The taylor series, evaluated at $x_0$, can be obtained by

$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$

and you should see this summation appearing naturally.

So your main task is to find the derivatives $f^{(n)}(x_0=0)$. Just differentiate $f(x)$ and plug in $x=0$. You should see the pattern. Note that $2n \choose n$$= \frac{2n!}{n!n!}$. Hence $f^{(n)}(x_0=0)$ should be equal to $\frac{2n!}{n!}$.

• Yes, I know what Taylor and Mclaurin series are but I don't have practice working with them. Where does the $1/4$ exponent on the first equation (function) come from? – Antoni Parellada Jul 5 '16 at 8:54
• Sry, had a typo. It should be $-1/2$. – MrYouMath Jul 5 '16 at 9:02
• $1+2 x+6 x^2+20 x^3+70 x^4+O\left(x^5\right)$ – Antoni Parellada Jul 5 '16 at 9:49
• $f(0)=1$; $f^1(x)=\frac{2}{(1-4x)^{3/2}}$, $f^1(0) = 2$, $2/1!=2$; $f^2(x)=\frac{12}{(1-4x)^{5/2}}$, $f^2(0) = 12$, $12/2!=6$; $f^3(x)=\frac{120}{(1-4x)^{7/2}}$, $f^3(0) = 120$, $120/3!=20$; $f^4(x)=\frac{1680}{(1-4x)^{9/2}}$, $f^4(0)=1680$, $1680/4!=70$. – Antoni Parellada Jul 5 '16 at 13:25