Sum 3.

$\begin{aligned} \underbrace{\sum_{n=1}^\infty \frac 1{n^2}}_{\text{all terms}} & = \underbrace{\sum_{n=0}^\infty \frac 1{(2n+1)^2}}_{\text{odd terms}} + \underbrace{\sum_{n=1}^\infty \frac 1{(2n)^2}}_{\text{even terms}} \\ \implies \sum_{n=0}^\infty \frac 1{(2n+1)^2} & = \sum_{n=1}^\infty \frac 1{n^2} - \sum_{n=1}^\infty \frac 1{(2n)^2} \\ & = \sum_{n=1}^\infty \frac 1{n^2} - \frac 14 \sum_{n=1}^\infty \frac 1{n^2} \\ & = \left(1-\frac 14\right)\sum_{n=1}^\infty \frac 1{n^2} \\ & = \frac 34 \sum_{n=1}^\infty \frac 1{n^2} \\ & = \frac 34 \times \frac {\pi^2}6 \\ & = \boxed{\dfrac {\pi^2}8} \end{aligned}$

$\displaystyle S_1 = \sum_{n=1}^{\infty} \dfrac{1}{n^2} = \dfrac{1}{1} + \dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{25} + \ldots = \displaystyle \dfrac{\pi^2}{6} \\ \displaystyle \text{Let, } \\ \displaystyle S_2 = \dfrac{1}{1} + \dfrac{1}{9} + \dfrac{1}{25} + \ldots = \sum_{n=1}^{\infty}\dfrac{1}{(2n+1)^2} \\ \displaystyle \implies S_2 = \dfrac{1}{1} + \color{#D61F06}{\dfrac{1}{4}} + \dfrac{1}{9} + \color{#D61F06}{\dfrac{1}{16}} + \dfrac{1}{25} + \ldots - \displaystyle \color{#D61F06}{\dfrac{1}{4}} - \color{#D61F06}{\dfrac{1}{16}} - \ldots \\ \displaystyle = \color{#3D99F6}{(\dfrac{1}{1} + \dfrac{1}{4} + \dfrac{1}{9} + \dfrac{1}{16} + \dfrac{1}{25} + \ldots)} - \color{#D61F06}{(\dfrac{1}{4} + \dfrac{1}{16} + \ldots)} \\ \displaystyle = \sum_{n=1}^{\infty}\dfrac{1}{n^2} - \sum_{n=1}^{\infty}\dfrac{1}{(2n)^2}\\ \displaystyle = \color{#3D99F6}{\sum_{n=1}^{\infty}\dfrac{1}{n^2}} - \color{#D61F06}{\dfrac{1}{4}}\color{#3D99F6}{\sum_{n=1}^{\infty}\dfrac{1}{n^2}} \\ \displaystyle = \dfrac{\pi^2}{6} - \dfrac{1}{4} \dfrac{\pi^2}{6} \\ \displaystyle = \dfrac{4\pi^2 - \pi^2}{24} \\ \displaystyle = \dfrac{3}{24} \pi^2 \\ \displaystyle = \dfrac{\pi^2}{8}$