Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 13"
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The sum <center><p><math>\sum_{k=1}^{2007} \arctan\left(\frac{1}{k^2+k+1}\right)</math></p></center> can be written in the form <math>\arctan\left(\frac{m}{n}\right)</math>, where <math>\gcd(m,n) = 1</math>. Compute the remainder when <math>m+n</math> is divided by 100. | The sum <center><p><math>\sum_{k=1}^{2007} \arctan\left(\frac{1}{k^2+k+1}\right)</math></p></center> can be written in the form <math>\arctan\left(\frac{m}{n}\right)</math>, where <math>\gcd(m,n) = 1</math>. Compute the remainder when <math>m+n</math> is divided by 100. | ||
==Solution== | ==Solution== | ||
| + | First let us develop a formula for sums of arctans. Let us say we want to find <math>\tan^{-1}{a} - \tan^{-1}{b}</math>, so we take the tangent of that and you get: | ||
| − | {{ | + | <math>\tan(\tan^{-1}{a} - \tan^{-1}{b}) = \frac{a - b}{1 + ab}</math>, now if we take the arctan of both sides we get: |
| + | <math>\tan^{-1}{a} - \tan^{-1}{b} = \tan^{-1}{\frac{a - b}{1 + ab}}</math> | ||
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| + | If we say that <math>x_k = \tan^{-1}{k}</math>, then we get | ||
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| + | <math>\tan{(x_{k+1} - x_k)} = \frac{\tan{x_{k+1}} - \tan{x_{k}}}{1 + \tan{x_{k+1}} \tan{x_{k}}} = \frac{1}{k^2 + k + 1}</math> | ||
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| + | Therefore the sum telescopes leaving us with <math>\sum_{k=1}^{2007}\tan^{-1}{\frac{1}{k^2 + k + 1}} = \tan^{-1}{2008} - \tan^{-1}{1} = \tan^{-1}{\frac{2007}{1 + 2008}} = \frac{2007}{2009}</math> and <math>2007 + 2009 \equiv \fbox{016}(\mod 100)</math> | ||
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| + | i don't know if my answer is right as there is no answer key for this test. | ||
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| + | ---- | ||
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| + | *[[Mock AIME 4 2006-2007 Problems/Problem 14| Next Problem]] | ||
| + | *[[Mock AIME 4 2006-2007 Problems/Problem 12| Previous Problem]] | ||
*[[Mock AIME 4 2006-2007 Problems]] | *[[Mock AIME 4 2006-2007 Problems]] | ||
Latest revision as of 12:50, 27 June 2008
Problem
The sum
can be written in the form 
, where 
. Compute the remainder when 
is divided by 100.
Solution
First let us develop a formula for sums of arctans. Let us say we want to find 
, so we take the tangent of that and you get:
, now if we take the arctan of both sides we get:
If we say that 
, then we get
Therefore the sum telescopes leaving us with 
and 
i don't know if my answer is right as there is no answer key for this test.
