AIME 2015 Problem 13

De Moiver's formula is widely used to calculate the summation however it is sometimes also calculated by using Telescopic summation method. Here its a version of the calculation using the telescopic method. $\begin{gathered} {a_n} = \sum\limits_{k = 1}^n {\sin k} \\ {\text{ }} = \frac{1}{2}\csc \left( {\frac{1}{2}} \right)\sum\limits_{k = 1}^n {\left( {2sin\left( {\frac{1}{2}} \right)\sin \left( {\frac{{2k}}{2}} \right)} \right)} \\ {\text{ }} = \frac{1}{2}\csc \left( {\frac{1}{2}} \right)\sum\limits_{k = 1}^n {\left( {\cos \left( {\frac{{2k - 1}}{2}} \right) - \cos \left( {\frac{{2k + 1}}{2}} \right)} \right)} \\ \end{gathered}$ $\begin{gathered} = \frac{1}{2}\csc \left( {\frac{1}{2}} \right)\left( {\cos \left( {\frac{1}{2}} \right) + \underbrace {\sum\limits_{k = 2}^n {\cos \left( {\frac{{2k - 1}}{2}} \right)} }_{ = \sum\limits_{k = 1}^{n - 1} {\cos \left( {\frac{{2k + 1}}{2}} \right)} } - \sum\limits_{k = 1}^{n - 1} {\cos \left( {\frac{{2k + 1}}{2}} \right)} - \cos \left( {\frac{{2n + 1}}{2}} \right)} \right){\text{ }} \\ = {\text{ }}\frac{1}{2}\csc \left( {\frac{1}{2}} \right)\left( {\cos \left( {\frac{1}{2}} \right) - \cos \left( {\frac{{2n + 1}}{2}} \right) + \sum\limits_{k = 1}^{n - 1} {\cos \left( {\frac{{2k + 1}}{2}} \right) - \sum\limits_{k = 1}^{n - 1} {\cos \left( {\frac{{2k + 1}}{2}} \right)} } } \right){\text{ }} \\ {\text{ = }}\frac{1}{2}\csc \left( {\frac{1}{2}} \right)\left( {\cos \left( {\frac{1}{2}} \right) - \cos \left( {\frac{{2n + 1}}{2}} \right)} \right) \\ \end{gathered}$ ${\text{now, }}{a_n} < 0{\text{ implies }}\frac{1}{2}\csc \left( {\frac{1}{2}} \right)\left( {\cos \left( {\frac{1}{2}} \right) - \cos \left( {\frac{{2n + 1}}{2}} \right)} \right) < 0$ $\begin{gathered} \Rightarrow \cos \left( {\frac{{2n + 1}}{2}} \right) > \cos \left( {\frac{1}{2}} \right){\text{ }} \\ \Rightarrow {\text{2co}}{{\text{s}}^2}\left( {\frac{{2n + 1}}{4}} \right) - 1 > {\text{2co}}{{\text{s}}^2}\left( {\frac{1}{4}} \right) - 1{\text{ }} \\ \Rightarrow {\text{co}}{{\text{s}}^2}\left( {\frac{{2n + 1}}{4}} \right) > {\text{co}}{{\text{s}}^2}\left( {\frac{1}{4}} \right){\text{ }} \\ \Rightarrow {\sec ^2}\left( {\frac{{2n + 1}}{4}} \right) < {\sec ^2}\left( {\frac{1}{4}} \right) \\ \Rightarrow 1 + {\tan ^2}\left( {\frac{{2n + 1}}{4}} \right) < 1 + {\tan ^2}\left( {\frac{1}{4}} \right){\text{ }} \\ \Rightarrow {\tan ^2}\left( {\frac{{2n + 1}}{4}} \right) < {\tan ^2}\left( {\frac{1}{4}} \right){\text{ }} \\ \Rightarrow \left| {\tan \left( {\frac{{2n + 1}}{4}} \right)} \right| < {\text{ }}\tan \left( {\frac{1}{4}} \right) \\ \Rightarrow - \tan \left( {\frac{1}{4}} \right) < \tan \left( {\frac{{2n + 1}}{4}} \right) < \tan \left( {\frac{1}{4}} \right) \\ \end{gathered}$ $\begin{gathered} \Rightarrow \tan \left( {m\pi - \frac{1}{4}} \right) < \tan \left( {\frac{{2n + 1}}{4}} \right) < \tan \left( {m\pi + \frac{1}{4}} \right),{\text{ becasue tan}}\left( x \right) \equiv \tan \left( {m\pi + x} \right){\text{ for all }}m \in \mathbb{Z} \\ \Rightarrow m\pi - \frac{1}{4} < \frac{{2n + 1}}{4} < m\pi + \frac{1}{4} \\ {\text{ }} \Rightarrow {\text{ }}2\pi m - 1 < n < 2\pi m \\ {\text{ }} \Rightarrow n \in (2\pi m - 1,2\pi m) \\ \end{gathered}$ $\begin{gathered} {\text{Here,m represents the number of first m cyclies of }} \\ {\text{the continuous function of }}\frac{1}{2}\csc \left( {\frac{1}{2}} \right)\left( {\cos \left( {\frac{1}{2}} \right) - \cos \left( {\frac{{2n + 1}}{2}} \right)} \right){\text{ }} \\ {\text{and also the numer of first m negative terms of the sequence because }} \\ {\text{there exists only 1 negative value of }}{a_n}{\text{ in each [}}2\pi (m - 1),2\pi m] \\ {\text{Now,the length of the interval (2m}}\pi {\text{ - 1,2m}}\pi {\text{) is 1 indicating that }} \\ {\text{only one integer value for n is possible in this interval }}{\text{.}} \\ {\text{For }}m = 100,{\text{ }}n \in (627.31...,628.31...){\text{ and since }}n \in \mathbb{Z},{\text{ }}n = \boxed{628}{\text{ }} \\ \end{gathered}$

$2*\pi=\dfrac{44} 7. \ \ So\ once \ in \ 2*\pi, there \ will \ be \ one \ -tive.\\$
So in 44 radians, there would be 7, negatives.
For any cycle, the index at the start of 7 step cycle is 6 * 6+1=37 less that the end index.
Next cycle starts after 7 more indices. So between cycles the similarly located indices are 44 apart.
From the first index of the cycle, next is at 6 more indices, till third.
Between 3rd and four there are 7 more indices.
From the forth index of the cycle, next is at 6 more indices, till the last-7th.
First cycle starts with first negative at index 6. Thus the 7th negative is at index 43.

So to have 98 negative, we need 14 * 44-1=615 as index.
For next cycle 7 more. That is for 99 negative, the index is 615+7=622.\
For 100 negatives, the index is 622+6= $\Large\ \ \ \color{#D61F06}{628}.$
In short negative appear as follows.
At (6-12-18--25-31-37-43)--(50-56-62--69-75-81-87)--(94 to 131).......(572 to 615)--622-628
In a cycle first at n, then n+6, n+12, n+19, n+25, n=31, n+37
So 7 negatives at 43,..... 14 at 43+44=87,..... 21 at 87+44=131,..... 7n at n * 44-1.
The diffrance $2*\pi-\dfrac{44} 7=-.002528$ would have effect for very large indices
and there will have to be a correction from the above method.

\text{The index f(x) for x appearances of negative value, for x<2251 is as under.}\\ \color{#3D99F6}{f(x)=\ \left \lfloo \dfrac{44*x} 7 \rigrt \rfloor} \ \ \ \ -\ \left \lfloor \dfrac x {161} \ \rigrt \rfloor}\ \ \left \lfloor \dfrac x {511} \ \rigrt \rfloor}\ -\ \left \lfloor \dfrac x {970} \ \rigrt \rfloor}\ -\ \left \lfloor \dfrac x {1355} \ \rigrt \rfloor}\ -\ \left \lfloor \dfrac x {1665} \ \rigrt \rfloor}\ }