Given $f(x) = \pi x - x^2 \hspace{1cm} (0, \pi)$

Half range sine series is given by

$b_{n} = \frac{2}{l}\int\limits_{0}^{\pi}f(x)sin nxdx$

$b_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}f(x)sinnxdx$

$b_{n} = \frac{2}{\pi}\int\limits_{0}^{\pi}(\pi x - x^2)sinnxdx$

$b_{n} = \frac{2}{\pi}\bigg[(\pi x - x^2)\bigg(\frac{-cosnx}{n}\bigg) - (\pi - 2x)\bigg(\frac{-sinx}{n}\bigg) + (-2) \bigg(\frac{cosnx}{n^3}\bigg)\bigg]_{0}^{\pi}$

$\therefore b_{n} = \frac{2}{\pi}\bigg[\bigg(0 - 0 + \bigg(\frac{-2cosn\pi}{n^3}\bigg) - 0 + 0 + \frac{2}{n^3}\bigg)\bigg]$

$b_{n} = \frac{2}{\pi}\bigg(\frac{-2cosn\pi}{n^3}\bigg) + \frac{2}{n^3}$

$b_{n} = \frac{2}{\pi}\bigg[\frac{4}{n^3}\bigg] = \frac{8}{\pi n^3} $ if n is odd

= 0 if n is even

$\therefore f(x) = \pi x - x^2 = \frac{8}{\pi}\bigg[\frac{1}{1^3}sinx + \frac{1}{3^3}sin3x + \frac{1}{5^3}sin5x.........\bigg]$

put x = $\frac{\pi}{2}$

$\therefore \pi * \frac{\pi}{2} - \frac{\pi^2}{4} = \frac{8}{\pi}\bigg[1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + .....\bigg]$

$\frac{\pi^2}{4} = \frac{8}{\pi}\bigg[1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} +.......\bigg]$

i.e.

$\frac{\pi^2}{32} = \bigg[1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + ......\bigg]$