Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
I have checked on google for $ \sum_{n=1}^\infty \frac{1}{n^2}$ and i do not suppose to know that is equal to $\frac{\pi^2}{6}$.
and I checked how The sum can be given explicitly of $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
and the result was show that needs the Fourier series and we haven't learned that yet.
So, I have no idea how to prove that equal to $\frac{3}{4}S$ without know the exactly the sum
Thanks.

Hint: Note that you get to this series by subtracting the even $n$ terms from the original.
Further, note that for any $n$,
$$
\frac{1}{(2n)^2}=\frac{1}{4}\cdot\frac{1}{n^2}.
$$
