Happy New Year!

A concise form is to evaluate the expression in the summation notation itself and using the concepts of the Riemann Zeta function, specifically the value $\zeta(2)=\dfrac{\pi^2}{6}$ which is the same as the Basel problem.

$S=\sum_{n=0}^\infty \left(\frac{1}{(2n+1)^2}\right)=\sum_{n=1}^\infty \left(\frac{1}{n^2}-\frac{1}{(2n)^2}\right)\\ \implies S=\sum_{n=1}^\infty \left(\frac{1}{n^2}-\frac{1}{4}\cdot \frac{1}{n^2}\right)$

Distributing the summation symbol inside the general term of the sequence, we have,

$S=\zeta(2)-\frac{1}{4}\zeta(2)=\frac{3}{4}\cdot \zeta(2)=\frac{3\pi^2}{24}=\boxed{\dfrac{\pi^2}{8}}$