高冪二項係数付き級数の関係式

${\displaystyle {\beta }_n:=\frac{(\genfrac{}{}{0ex}{}{2n}{n})}{2^{2n}}}$
気が向き次第順次追加します

$$\begin{array}{rl}& \frac{{\pi }^2}{4}\sum _{0\le n}{\beta }_n^5(2\beta (2)+\sum _{m=1}^n\frac{1}{(2m{)}^2{\beta }_m^2})+\frac{2}{\pi }\sum _{0<n}\frac{1}{(2n{)}^5{\beta }_n^5}{\left(\sum _{m=0}^{n-1}{\beta }_m^2\right)}^2\\ & =\sum _{0<n}\frac{1}{(2n{)}^2{\beta }_n}\sum _{m=0}^{n-1}{\beta }_m^5(2\beta (2)+\sum _{k=1}^m\frac{1}{(2k{)}^2{\beta }_k^2})+\sum _{0\le n}\frac{1}{(2n+1{)}^2{\beta }_n}\sum _{m=0}^n{\beta }_m^5(2\beta (2)+\sum _{k=1}^m\frac{1}{(2k{)}^2{\beta }_k^2})\end{array}$$

$$\begin{array}{rl}& \frac{{\pi }^2}{4}\sum _{0\le n}{\beta }_n^5(2\beta (2)+\sum _{m=1}^n\frac{1}{(2m{)}^2{\beta }_m^2})+\sum _{0\le n}{\beta }_n^5{(2\beta (2)+\sum _{m=1}^n\frac{1}{(2m{)}^2{\beta }_m^2})}^2\\ =& \frac{\pi }{2}\sum _{0\le n}{\beta }_n^5(\frac{7}{2}\zeta (3)+\sum _{m=1}^n\frac{4m-1}{(2m{)}^4{\beta }_m^4})+\sum _{0<n}\frac{1}{(2n{)}^5{\beta }_n^5}\left(\sum _{m=0}^{n-1}{\beta }_m^2\right)\sum _{m=0}^{n-1}{\beta }_m^4(4m+1)\end{array}$$

$$\begin{array}{rl}\pi \sum _{0<n}\frac{1}{(2n{)}^4{\beta }_n^4}{\left(\sum _{m=0}^{n-1}{\beta }_m^2\right)}^3+& \sum _{0<n}\frac{1}{(2n{)}^4{\beta }_n^4}\left(\sum _{m=0}^{n-1}{\beta }_m^2\right)\sum _{m=0}^{n-1}{\beta }_m^2(\ln 2+\sum _{k=0}^{m-1}\frac{1}{2k+1})\\ =\pi & \sum _{0<n}\frac{1}{(2n{)}^4{\beta }_n^4}(\sum _{m=0}^{n-1}(4m+1){\beta }_m^4)\sum _{m=0}^{n-1}{\beta }_m^2(2\ln 2+\sum _{k=1}^m\frac{1}{2k})\end{array}$$