A Series Problem Based On Trigonometric Function.

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A Series Problem Based On Trigonometric Function.

image containing math problem statement

A series problem based on trigonometric function. Can you spot the solution here? Try looking at it through a telescope.

Problem contributed by anonymous user

We can evaluate the LHS of the equation as follows:

$\tan \left( \sum \limits_{n=1}^{100} \tan^{-1}\left( \dfrac{1}{n^2 + n + 1} \right) \right)$

$= \tan \left( \sum \limits_{n=1}^{100} \tan^{-1}\left( \dfrac{1}{n(n + 1) + 1} \right) \right)$

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$= \tan \left( \sum \limits_{n=1}^{100} \tan^{-1}\left( \dfrac{(n+1) - n}{1 + n(n + 1)} \right) \right)$

$= \tan \left( \sum \limits_{n=1}^{100} \left( \tan^{-1} (n+1) - \tan^{-1} (n) \right) \right)$

$= \tan \left( \left( \cancel{\tan^{-1}2} - \tan^{-1}1 \right) + \left( \cancel{\tan^{-1}3} - \cancel{\tan^{-1}2} \right) \right.$

$\left. + \left( \cancel{\tan^{-1}3} - \cancel{\tan^{-1}4} \right) + ... + \left( \tan^{-1}101 - \cancel{\tan^{-1}100} \right)\right)$

$= \tan \left( \tan^{-1}101 - \tan^{-1}1 \right)$

$= \tan \left( \tan^{-1} \left( \dfrac{101 - 1}{1 + 101 \cdot 1} \right) \right)$

$= \dfrac{50}{51}$

Therefore, $a = 50$, $b = 51$ and $(a + b) = 101$