Question

It is given that ${\sum }_{r=1}^{\infty }\frac{1}{(2r-1{)}^2}=\frac{{\pi }^2}{8}$ and ${\sum }_{r=1}^{\infty }\frac{1}{r^2}$ is equal to $\frac{{\pi }^2}{2k}$,then k =___

Answer 1

$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+.....upto\infty =\frac{{\pi }^2}{8}Letx=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....\infty =(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....\infty )+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+....\infty )x=\frac{{\pi }^2}{8}+\frac{1}{4}xx=\frac{{\pi }^2}{6}Now2k=6\Rightarrow k=3$