Thread: An infinite series

We know that :

$\displaystyle \sum_{k=0}^{\infty} ( \frac{x}{4} )^k \binom{2k}{k} = \frac{1}{ \sqrt{ 1 - x}} , |x| < 1$

I'd like to ask what about

$\displaystyle \sum_{k=0}^{\infty} ( \frac{x}{16} )^k \binom{4k}{2k} , |x| < 1 ~~ $ ?

Originally Posted by simplependulum

We know that :

$\displaystyle \sum_{k=0}^{\infty} ( \frac{x}{4} )^k \binom{2k}{k} = \frac{1}{ \sqrt{ 1 - x}} , |x| < 1$

I'd like to ask what about

$\displaystyle \sum_{k=0}^{\infty} ( \frac{x}{16} )^k \binom{4k}{2k} , |x| < 1 ~~ $ ?

If

$\displaystyle \frac{1}{ \sqrt{ 1 - x}} = \sum_{k=0}^{\infty} ( \frac{x}{4} )^k \binom{2k}{k} = 1 + \binom{2}{1} \frac{x}{4} + \binom{4}{2} \left(\frac{x}{4}\right)^2 + \binom{6}{3} \left(\frac{x}{4}\right)^3 + \binom{8}{4} \left(\frac{x}{4}\right)^4 \cdots $

then

$\displaystyle \frac{1}{ \sqrt{ 1 + x}} = 1 - \binom{2}{1} \frac{x}{4} + \binom{4}{2} \left(\frac{x}{4}\right)^2 - \binom{6}{3} \left(\frac{x}{4}\right)^3 + \binom{8}{4} \left(\frac{x}{4}\right)^4 \cdots $

then

$\displaystyle \frac{1}{2}\left( \frac{1}{ \sqrt{ 1 - x}}+ \frac{1}{\sqrt{ 1 + x}}\right) = 1 + \binom{4}{2} \left(\frac{x}{4}\right)^2 + \binom{8}{4} \left(\frac{x}{4}\right)^4 \cdots $

then replace $\displaystyle x$ with $\displaystyle \sqrt{x}$.