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Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $

My approach is as follow

$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \right)} $

$ \Rightarrow \sum\limits_{k = 0}^n {\left( 1 \right)} - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \Rightarrow \left( {n + 1} \right) - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} $

$\sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} = \sum\limits_{k = 0}^n {\cos \left( {\frac{{2\pi }}{{n + 2}} + \frac{2}{{n + 2}}k} \right)} ;a = \frac{{2\pi }}{{n + 2}};d = \frac{2}{{n + 2}}$

From the website I got the following formula but the summation of the series is from $0$ to $n-1$.

$\sum\limits_{k = 0}^{n - 1} {\cos \left( {a + kd} \right)} = \frac{{\sin \left( {\frac{{nd}}{2}} \right)}}{{\sin \left( {\frac{d}{2}} \right)}} \times \cos \left( {a + \frac{{\left( {n - 1} \right)d}}{2}} \right)$

Where as in the question it is from $0$ to $n$, a total of $n+1$ terms. How do I proceed

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2 Answers 2

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$\dfrac{\sin\frac{(n+1)d}{2}}{\sin \frac{d}{2}} \cdot \cos(a + \dfrac{nd}{2})$

For summation from $0$ to $n$
That will work

A very basic proof if you want would be substitute n = t -1 and then using the summation formula for cosine.

Proof of summation formula of cosine and sine up to n-1 terms

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Move the last term of the sum out of the sum. The remaining sum will match your reference. $$\sum_{k=0}^n f(k) = \left( \sum_{k=0}^{n-1} f(k) \right) + f(n) \text{.}$$

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