Mock AIME 3 Pre 2005 Problems/Problem 15

Problem

Let $\Omega$ denote the value of the sum

$\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)$

The value of $\tan\left(\Omega\right)$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Compute $m + n$ .

Solution

Let $x := k^2 + k + 1$ $a_k := \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}} \right )$

Factoring the radicand, we have $a_k = \cos^{-1}\left(\frac{x}{\sqrt{x^2+1}} \right )$ The fraction looks remarkably apt for a trigonometric substitution; namely, define $\theta$ such that $x = \tan{\theta}$ . Then the RHS becomes $a_k = \cos^{-1}\left(\frac{\tan{\theta}}{\sec{\theta}} \right ) = \cos^{-1}{(\sin{\theta})}$ But $\cos{(\pi/2-\theta)} = \sin{\theta}$ Therefore, $a_k = \pi/2 - \theta = \pi/2 - \tan^{-1}{x}$ This gives us $\tan{a_k} = \tan{(\pi/2 - \tan^{-1}{x})}$ $= \cot{(\tan^{-1}{x})} = \dfrac{1}{\tan{(\tan^{-1}{x})} }$ $= \dfrac{1}{x} = \dfrac{1}{k^2+k+1}$ So now $a_k = \tan^{-1}{\left( \dfrac{1}{k^2+k+1} \right) }$ $= \tan^{-1}{ \left( \dfrac{ (k+1) + (-k) }{ 1 - (-k)(k+1) } \right ) }$ $= \tan^{-1}{(k+1)} - \tan^{-1}{k}$ When we sum $a_k$ , this sum now telescopes: $\Omega = \sum_{k=1}^{40} a_k$ $= \sum_{k=1}^{40} \tan^{-1}{(k+1)} - \tan^{-1}{k}$ $= \tan^{-1}{41} - \tan^{-1}{1}$ Therefore, the required value $\tan{\Omega} = \tan{(\tan^{-1}{41} - \tan^{-1}{1})}$ $= \dfrac{ 41 - 1}{1 + 40 \cdot 1 }$ $= \dfrac{40}{41}$ giving us the desired answer of $\boxed{81}$ .

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Mock AIME 3 Pre 2005 Problems/Problem 15

Problem

Solution

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