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Mock AIME 3 Pre 2005 Problems/Problem 15

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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 \[\tan{\Omega} = \tan{(\tan^{-1}{41} - \tan^{-1}{1})}\] \[= \dfrac{ 41 - 1}{1 + 41 \cdot 1 }\] \[= \dfrac{40}{42}\] giving us the desired answer of $20+21=\boxed{41}$.

See Also

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by
Problem 14
Followed by
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