It is given that,

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

Simplifying the above expression we get,

$\sum _{k=1}^n{\tan }^{-1}\left(\frac{(k+1)-k}{1+k(k+1)}\right)={\tan }^{-1}\theta$

$\Rightarrow \sum _{k=1}^n\left\{{\tan }^{-1}(k+1)-{\tan }^{-1}k\right\}={\tan }^{-1}\theta$

$\left\{\begin{array}{l}{\tan }^{-1}2-{\tan }^{-1}1+{\tan }^{-1}3-{\tan }^{-1}2+{\tan }^{-1}4-\\ {\tan }^{-1}3+\dots +{\tan }^{-1}(n+1)-{\tan }^{-1}n\end{array}\right\}={\tan }^{-1}\theta$

$\Rightarrow {\tan }^{-1}(n+1)-{\tan }^{-1}1={\tan }^{-1}\theta$

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

$\Rightarrow \frac{n}{2+n}=\theta$