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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

MyMolecules
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Martin.s
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  • Your equation $\sin(\pi/(m+1))=\sin(k\pi/(m+1))$ is not generally true. Also, it is not clear for me how you are applying the base case. – Jam Sep 17 '23 at 18:33
  • @Jam however, if $k=m$, then it becomes true. – Kamal Saleh Sep 17 '23 at 18:50
  • @Eric If the question is to find out what is wrong, then it doesn't. This is most likely because the OP stops at a false conclusion. But if they are looking for a solution to the problem, then it would be a duplicate. Martin, please clarify the exact problem you are finding the answer of. – Kamal Saleh Sep 17 '23 at 19:32
  • @Eric Yes I don't know this question was already on mse – Martin.s Sep 17 '23 at 20:16

1 Answers1

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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

Kamal Saleh
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