Question

If $\sum _{n=1}^{100}{\tan }^{-1}(n^2+n+1)=\frac{a\pi }{b}-{\tan }^{-1}c,$ where $a,b,c\in N$ and $a$ and $b$ are relatively prime, then the value of $\frac{a-c}{b}$ is

Answer 1

${\tan }^{-1}(n^2+n+1)$
$={\tan }^{-1}\left(\frac{n(n+1)+1}{(n+1)-n}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)$
$=\frac{\pi }{2}-\left({\tan }^{-1}\right(n+1)-{\tan }^{-1}n)$

$\therefore \sum _{n=1}^{100}{\tan }^{-1}(n^2+n+1)$
$=100\left(\frac{\pi }{2}\right)-({\tan }^{-1}101-{\tan }^{-1}1)=50\pi +\frac{\pi }{4}-{\tan }^{-1}101$
$=\frac{201\pi }{4}-{\tan }^{-1}\left(101\right)\therefore \frac{a-c}{b}=\frac{201-101}{4}=25$