↑ stuart clark:
Hi,
Skrytý text:the sum can be written in a form
[mathjax] {\displaystyle \frac{8}{9}\sum_{k=2}^{\infty}\binom{2k}{k}(2k+1)\left(\frac{3}{16}\right)^{k} }[/mathjax]From what i wrote previously
here it follows that there holds
[mathjax] {\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}x^k = \frac{1}{\sqrt{1-4x}}=:f(x) }[/mathjax].
Consequently for the function
[mathjax] {\displaystyle g(x)=xf(x^2)=\sum_{k=0}^{\infty}\binom{2k}{k}x^{2k+1} }[/mathjax] it holds
[mathjax] {\displaystyle g'(x) = \sum_{k=0}^{\infty}\binom{2k}{k}(2k+1)x^{2k} = \left(\frac{x}{\sqrt{1-4x^2}}\right)' = \frac{1}{(1-4x^2)^{3/2}} } [/mathjax]Hence
[mathjax] {\displaystyle \frac{8}{9}\sum_{k=2}^{\infty}\binom{2k}{k}(2k+1)\left(\frac{3}{16}\right)^{k} = \frac{8}{9}\left[g'\left(\frac{\sqrt{3}}{4}\right) - 1 - 6\cdot\frac{3}{16} \right] = \frac{8}{9}\left[ \frac{1}{(1-3/4)^{3/2}} - 1 - \frac{9}{8} \right] = \frac{8}{9}\left[8-1-\frac{9}{8}\right]=\frac{47}{9} } [/mathjax]