# Mock AIME 4 2006-2007 Problems/Problem 13

## Problem

The sum $\sum_{k=1}^{2007} \arctan\left(\frac{1}{k^2+k+1}\right)$

can be written in the form $\arctan\left(\frac{m}{n}\right)$, where $\gcd(m,n) = 1$. Compute the remainder when $m+n$ is divided by 100.

## Solution

First let us develop a formula for sums of arctans. Let us say we want to find $\tan^{-1}{a} - \tan^{-1}{b}$, so we take the tangent of that and you get: $\tan(\tan^{-1}{a} - \tan^{-1}{b}) = \frac{a - b}{1 + ab}$, now if we take the arctan of both sides we get: $\tan^{-1}{a} - \tan^{-1}{b} = \tan^{-1}{\frac{a - b}{1 + ab}}$

If we say that $x_k = \tan^{-1}{k}$, then we get $\tan{(x_{k+1} - x_k)} = \frac{\tan{x_{k+1}} - \tan{x_{k}}}{1 + \tan{x_{k+1}} \tan{x_{k}}} = \frac{1}{k^2 + k + 1}$

Therefore the sum telescopes leaving us with $\sum_{k=1}^{2007}\tan^{-1}{\frac{1}{k^2 + k + 1}} = \tan^{-1}{2008} - \tan^{-1}{1} = \tan^{-1}{\frac{2007}{1 + 2008}} = \frac{2007}{2009}$ and $2007 + 2009 \equiv \fbox{016}(\mod 100)$

i don't know if my answer is right as there is no answer key for this test.