Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 15"
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<cmath>\Omega = \sum_{k=1}^{40} a_k </cmath> | <cmath>\Omega = \sum_{k=1}^{40} a_k </cmath> | ||
<cmath> = \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k} </cmath> | <cmath> = \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k} </cmath> | ||
− | <cmath> = \tan^{-1}{ | + | <cmath> = \tan^{-1}{41} - \tan^{-1}{1} </cmath> |
Therefore, the required value | Therefore, the required value | ||
− | <cmath> \tan{\Omega} = \tan{(\tan^{-1}{ | + | <cmath> \tan{\Omega} = \tan{(\tan^{-1}{41} - \tan^{-1}{1})} </cmath> |
− | <cmath> = \dfrac{ | + | <cmath> = \dfrac{ 41 - 1}{1 + 41 \cdot 1 } </cmath> |
− | <cmath> = \dfrac{ | + | <cmath> = \dfrac{40}{42} </cmath> |
− | giving us the desired answer of <math>\boxed{ | + | giving us the desired answer of <math>20+21=\boxed{41}</math>. |
− | ==See | + | |
+ | ==See Also== | ||
+ | {{Mock AIME box|year=Pre 2005|n=3|num-b=14|after=Last Question}} |
Latest revision as of 09:38, 4 April 2012
Problem
Let denote the value of the sum
The value of can be expressed as , where and are relatively prime positive integers. Compute .
Solution
Let
Factoring the radicand, we have The fraction looks remarkably apt for a trigonometric substitution; namely, define such that . Then the RHS becomes But Therefore, This gives us So now When we sum , this sum now telescopes: Therefore, the required value giving us the desired answer of .
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |