Difference between revisions of "2008 AIME II Problems/Problem 8"
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By the [[trigonometric identity|product-to-sum identities]], we have that <math>2\cos a \sin b = \sin (a+b) - \sin (a-b)</math>. Therefore, this reduces to a [[telescope|telescoping series]]: | By the [[trigonometric identity|product-to-sum identities]], we have that <math>2\cos a \sin b = \sin (a+b) - \sin (a-b)</math>. Therefore, this reduces to a [[telescope|telescoping series]]: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} \sin(k(k+1)a) - \sin((k-1)ka)\\ | + | \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ |
| − | &= -\sin(0) + \sin(2a) | + | &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\ |
| − | + | &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) | |
| − | |||
| − | |||
| − | &= \sin(n(n+1)a) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | Thus, we need <math>\sin \left(\frac{n(n+1)\pi}{2008}\right)</math> to be an integer; this can be only <math>\{-1,0,1\}</math>, which occur when <math>2 \cdot \frac{n(n+1)}{2008}</math> is an integer. Thus <math>1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1</math>. It easily follows that <math>n = \boxed{251}</math> is the smallest such integer. | + | Thus, we need <math>\sin \left(\frac{n(n+1)\pi}{2008}\right)</math> to be an integer; this can be only <math>\{-1,0,1\}</math>, which occur when <math>2 \cdot \frac{n(n+1)}{2008}</math> is an integer. Thus <math>1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1</math>. It easily follows that <math>n = \boxed{251}</math> is the smallest such integer. |
== See also == | == See also == | ||
Revision as of 16:56, 15 April 2008
Problem
Let 
. Find the smallest positive integer 
such that
![\[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\]](http://latex.artofproblemsolving.com/1/3/0/130591afb17e8bdcc9d7334e4fcc26d57a8f65bd.png)
is an integer.
Solution
By the product-to-sum identities, we have that 
. Therefore, this reduces to a telescoping series:
![\begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\ &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) \end{align*}](http://latex.artofproblemsolving.com/4/1/7/417655839b4578551543e9697b058fe3d9c42e97.png)
Thus, we need 
to be an integer; this can be only 
, which occur when 
is an integer. Thus 
. It easily follows that 
is the smallest such integer.
See also
| 2008 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
