For $n = 2$:
- Left-hand side (LHS): $\sum_{k=1}^{1} \frac{1}{\sin^2\left(\frac{\pi k}{2}\right)} = \frac{1}{\sin^2\left(\frac{\pi}{2}\right)} = 1$
- Right-hand side (RHS): $\frac{1}{3}\left(2^2 - 1\right) = \frac{1}{3}(4-1) = 1$
Now, assume that the identity holds for some positive integer m, i.e.,
$\sum_{k=1}^{m-1} \frac{1}{\sin^2\left(\frac{\pi k}{m}\right)} = \frac{1}{3}\left(m^2 - 1\right)$
We need to show that it holds for m+1:
$\sum_{k=1}^{m} \frac{1}{\sin^2\left(\frac{\pi k}{m+1}\right)}$
We can use the identity $\sin\left(\frac{\pi}{m+1}\right) = \sin\left(\frac{\pi k}{m+1}\right)$
$\sum_{k=1}^{m} \frac{1}{\sin^2\left(\frac{\pi k}{m+1}\right)} = \sum_{k=1}^{m} \frac{1}{\sin^2\left(\frac{\pi}{m+1}\cdot k\right)}$Now, you can use the assumption for m:$= \frac{1}{3}\left(m^2 - 1\right) + \frac{1}{\sin^2\left(\frac{\pi}{m+1}\cdot (m+1)\right)}$$= \frac{1}{3}\left(m^2 - 1\right) + \frac{1}{\sin^2\left(\pi\right)}$
Any help will be appreciated Tysm
Your mistake here is that you thought $$\sum_{k=1}^{m-1}\frac1{\sin^2\left(\frac{k\pi}{m+1}\right)}=\sum_{k=1}^{m-1}\frac1{\sin^2\left(\frac{k\pi}{m}\right)}$$Which isn't true. If you want to find a proof of your claim, see here