If $f\left(n\right)=\tan \left[{\tan }^{-1}\frac{1}{1+2}+{\tan }^{-1}\frac{1}{1+6}+{\tan }^{-1}\frac{1}{1+12}+...+{\tan }^{-1}\frac{1}{1+n(n+1)}\right]$
then $f\left(2021\right)=$

Given $f\left(n\right)=\tan \left[{\tan }^{-1}\frac{1}{1+2}+{\tan }^{-1}\frac{1}{1+6}+{\tan }^{-1}\frac{1}{1+12}+...+{\tan }^{-1}\frac{1}{1+n(n+1)}\right]$

$\Rightarrow f\left(n\right)=\tan \left[{\tan }^{-1}\left(\frac{2-1}{1+2\times 1}\right)+{\tan }^{-1}\left(\frac{3-2}{1+3\times 2}\right)+{\tan }^{-1}\left(\frac{4-3}{1+4\times 3}\right)+...+{\tan }^{-1}\left(\frac{\left(n+1\right)-n}{1+n(n+1)}\right)\right]$

Using ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right),$

$\Rightarrow f\left(n\right)=\tan \left[{\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(4\right)-{\tan }^{-1}\left(3\right)+...+{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}\left(n\right)\right]$

$\Rightarrow f\left(n\right)=\tan \left[{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}\left(1\right)\right]$

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

$\Rightarrow f\left(n\right)=\tan \left[{\tan }^{-1}\left(\frac{n}{n+2}\right)\right]=\frac{n}{n+2}.$

$\Rightarrow f\left(2021\right)=\frac{2021}{2023}.$