Mathematics - Sequence & Series Question with Solution | TestHub

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots ..\infty =\frac{{\pi }^4}{90}....(i)$
$\begin{array}{rl}& \beta =\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\dots ,\\ & =\frac{1}{16}\left[\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\dots ..\right],\end{array}$
$=\frac{1}{16}\times \frac{{\pi }^4}{90}$
using (ii) ________...(ii)
$\begin{array}{rl}& \alpha =\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\dots ..\infty \\ & \left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\dots ..\right)\\ & -\left(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\dots ..\right)\end{array}$
$\alpha =\frac{{\pi }^4}{90}-\frac{1}{16}\times \frac{{\pi }^4}{90}[$ using (i) and (ii)]
$\begin{array}{rl}& \alpha =\frac{16-1}{16\times 90}\times {\pi }^4=\frac{15}{16\times 90}{\pi }^4=\frac{{\pi }^4}{96}\\ & \therefore \frac{\alpha }{\beta }=\frac{\frac{{\pi }^4}{96}}{\frac{{\pi }^4}{16\times 90}}=\frac{16\times 90}{96}=15\end{array}$