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Math Help - An infinite series

  1. #1
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    An infinite series

    We know that :

     \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

      \sum_{k=0}^{\infty} ( \frac{x}{16} )^k \binom{4k}{2k}  , |x| < 1  ~~ ?
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    We know that :

     \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

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

     \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

     \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

     \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 x with \sqrt{x}.
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