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 + 40 \cdot 1 } </cmath> |
| − | <cmath> = \dfrac{ | + | <cmath> = \dfrac{40}{41} </cmath> |
| − | giving us the desired answer of <math>\boxed{ | + | giving us the desired answer of <math>\boxed{81}</math>. |
| + | |||
==See also== | ==See also== | ||
Revision as of 16:41, 16 August 2010
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 ![\[x := k^2 + k + 1\]](http://latex.artofproblemsolving.com/b/3/a/b3a5dbe6e4c1cabeb2676cf9ec5e0774195eff8b.png)
![\[a_k := \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}} \right )\]](http://latex.artofproblemsolving.com/e/c/d/ecd8f9efe5263252b8fb69b68fb71da4f6d761ce.png)
Factoring the radicand, we have
![\[a_k = \cos^{-1}\left(\frac{x}{\sqrt{x^2+1}} \right )\]](http://latex.artofproblemsolving.com/4/1/2/41277565bf5d8c22ae06d1fb9899acf39839ee36.png)
The fraction looks remarkably apt for a trigonometric substitution; namely, define 
such that 
. Then the RHS becomes
![\[a_k = \cos^{-1}\left(\frac{\tan{\theta}}{\sec{\theta}} \right ) = \cos^{-1}{(\sin{\theta})}\]](http://latex.artofproblemsolving.com/f/7/1/f71054771c37d3384eb57ae8866b5620ba8ea0e7.png)
But ![\[\cos{(\pi/2-\theta)} = \sin{\theta}\]](http://latex.artofproblemsolving.com/6/0/c/60cceebd51d69ec31f8a394517746b54437b1b83.png)
Therefore,
![\[a_k = \pi/2 - \theta = \pi/2 - \tan^{-1}{x}\]](http://latex.artofproblemsolving.com/0/d/7/0d7cb58b222e40c73946da71a6704977023d16e3.png)
This gives us
![\[\tan{a_k} = \tan{(\pi/2 - \tan^{-1}{x})}\]](http://latex.artofproblemsolving.com/c/a/3/ca305dbdac1e9c3d40a9b0514f103be0e1c33c64.png)
![\[= \cot{(\tan^{-1}{x})} = \dfrac{1}{\tan{(\tan^{-1}{x})} }\]](http://latex.artofproblemsolving.com/1/7/8/178ad8ace71c8fef0cfdd15c8e53a6b715213df2.png)
![\[= \dfrac{1}{x} = \dfrac{1}{k^2+k+1}\]](http://latex.artofproblemsolving.com/a/e/2/ae2fd08b387d5a1af081c9f542a79379bf4d5e20.png)
So now
![\[a_k = \tan^{-1}{\left( \dfrac{1}{k^2+k+1} \right) }\]](http://latex.artofproblemsolving.com/a/d/6/ad604c93b5bd301217636d312b6898e366461f41.png)
![\[= \tan^{-1}{ \left( \dfrac{ (k+1) + (-k) }{ 1 - (-k)(k+1) } \right ) }\]](http://latex.artofproblemsolving.com/0/6/8/06825b2f63f44bba692c6ab669db2a2a74267222.png)
![\[= \tan^{-1}{(k+1)} - \tan^{-1}{k}\]](http://latex.artofproblemsolving.com/2/2/d/22d8ae36de25efba9a2aed023d8127daec200c62.png)
When we sum 
, this sum now telescopes:
![\[\Omega = \sum_{k=1}^{40} a_k\]](http://latex.artofproblemsolving.com/f/d/e/fdea3ec619d95c9321af1e14994ea599e2679e4f.png)
![\[= \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k}\]](http://latex.artofproblemsolving.com/1/c/b/1cb518e647afda88c64bf03c116b601c26c8ceea.png)
![\[= \tan^{-1}{41} - \tan^{-1}{1}\]](http://latex.artofproblemsolving.com/c/2/8/c287502c69d1f76e6197038ddbb6e5b3579f85db.png)
Therefore, the required value
![\[\tan{\Omega} = \tan{(\tan^{-1}{41} - \tan^{-1}{1})}\]](http://latex.artofproblemsolving.com/1/c/a/1cab3ce310698162e8875eaaa00d72461ac43507.png)
![\[= \dfrac{ 41 - 1}{1 + 40 \cdot 1 }\]](http://latex.artofproblemsolving.com/4/e/4/4e4f928f558aa75f7f982b30599ab1fad1ff4899.png)
![\[= \dfrac{40}{41}\]](http://latex.artofproblemsolving.com/9/2/a/92a1d0a1f98ac150b7acce6994c1b2d61b2bbd86.png)
giving us the desired answer of 
.
