If S is the sum of the first 10 terms of the series ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+{\tan }^{-1}\left(\frac{1}{21}\right)+\mathrm{.....},$ then $\tan S$ equal to 

General term of sequence 

$t_n=a{n}^2+bn+c$

$\begin{array}{l}{t}_1=3=a+b+c\\ t_2=7=4a+2b+c\\ t_3=13=9a+3b+c\end{array}$

will be of the type So, we take $t_2-t_1$ and $t_3-t_2$

$\Rightarrow 3a+b=4and5a+b=6$$\Rightarrow 2a=2ora=1$

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

So, sum of first 10 terms will be $\sum _{n=1}^{10}\left({\tan }^{-1}\left(n+1\right)-{\tan }^{-1}n\right)$= ${\tan }^{-1}11-{\tan }^{-1}1={\tan }^{-1}\left(\frac{11-1}{1+\left(11\right)1}\right)={\tan }^{-1}\frac{5}{6}=S$ $\Rightarrow \tan S=\frac{5}{6} \left(\sin ce {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)\right)$