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Difference between revisions of "2015 AIME II Problems/Problem 13"

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==Solution==
 
==Solution==
  
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If <math>n = 1</math>, <math>a_n = sin(1) > 0</math>.  Then if <math>n</math> satisfies <math>a_n < 0</math>, <math>n \ge 2</math>, and
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If <math>n = 1</math>, <math>a_n = \sin(1) > 0</math>.  Then if <math>n</math> satisfies <math>a_n < 0</math>, <math>n \ge 2</math>, and
 
<cmath>a_n =\cfrac{1}{\sin{1}} \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath>
 
<cmath>a_n =\cfrac{1}{\sin{1}} \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].</cmath>
 
Since <math>\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos(\cfrac{1}{2})\cos(\cfrac{1}{2})</math> and <math>\cos(n) + \cos(n + 1) = 2\cos(n + \cfrac{1}{2})\cos(\cfrac{1}{2})</math>, <math>b_n</math> is negative if and only if <math>\cos(\cfrac{1}{2}) < \cos(n + \cfrac{1}{2})</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 6.28318 \rfloor = \boxed{628}</math>.
 
Since <math>\sin 1</math> is positive, it does not affect the sign of <math>a_n</math>.  Let <math>b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)</math>.  Now since <math>\cos(0) + \cos(1) = 2\cos(\cfrac{1}{2})\cos(\cfrac{1}{2})</math> and <math>\cos(n) + \cos(n + 1) = 2\cos(n + \cfrac{1}{2})\cos(\cfrac{1}{2})</math>, <math>b_n</math> is negative if and only if <math>\cos(\cfrac{1}{2}) < \cos(n + \cfrac{1}{2})</math>, or when <math>n \in [2k\pi - 1, 2k\pi]</math>.  Since <math>\pi</math> is irrational, there is always only one integer in the range, so there are values of <math>n</math> such that <math>a_n < 0</math> at <math>2\pi, 4\pi, \cdots</math>.  Then the hundredth such value will be when <math>k = 100</math> and <math>n = \lfloor 200\pi \rfloor = \lfloor 6.28318 \rfloor = \boxed{628}</math>.

Revision as of 23:24, 26 March 2015

Problem

Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$, where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$.

Solution

If $n = 1$, $a_n = \sin(1) > 0$. Then if $n$ satisfies $a_n < 0$, $n \ge 2$, and \[a_n =\cfrac{1}{\sin{1}} \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $\sin 1$ is positive, it does not affect the sign of $a_n$. Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$. Now since $\cos(0) + \cos(1) = 2\cos(\cfrac{1}{2})\cos(\cfrac{1}{2})$ and $\cos(n) + \cos(n + 1) = 2\cos(n + \cfrac{1}{2})\cos(\cfrac{1}{2})$, $b_n$ is negative if and only if $\cos(\cfrac{1}{2}) < \cos(n + \cfrac{1}{2})$, or when $n \in [2k\pi - 1, 2k\pi]$. Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$. Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 6.28318 \rfloor = \boxed{628}$.

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