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