Question

Prove that ${\sum }_{\lambda =1}^n{tan}^{-1}\left[\frac{2\lambda }{2+{\lambda }^2+{\lambda }^4}\right]={tan}^{-1}(n^2+n+1)-\frac{\pi }{4}$

Answer 1

We know that ${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$

$=\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.

Hence $2+{\lambda }^2+{\lambda }^4=1+({\lambda }^4+{\lambda }^2+1)$
$=1+\left[\right({\lambda }^2+\lambda +1\left)\right({\lambda }^2-\lambda +1\left)\right]$
and $2\lambda =({\lambda }^2+\lambda +1)-({\lambda }^2-\lambda +1)$ etc.
$\therefore$ $T_n={tan}^{-1}({\lambda }^2+\lambda +1)-{tan}^{-1}({\lambda }^2-\lambda +1)$
Now put $\lambda =1,2,3,....n$ and add.
$\therefore$ $S_n={tan}^{-1}(n^2+n+1)-{tan}^{-1}1={tan}^{-1}(n^2+n+1)-\frac{\pi }{4}$