Given expression $={\text{tan}}^{\mathrm{-1 }}2+{\text{tan}}^{\mathrm{-1 }}\frac{2}{9}+{\text{tan}}^{\mathrm{-1 }}\frac{2}{25}+{\text{tan}}^{\mathrm{-1 }}\frac{2}{49}+\text{........}$

$\text{General term}=\frac{2}{{\left(2n-1\right)}^2}=\frac{2}{4n^2-4n+1}=\frac{2}{1+4n\left(n-1\right)}=\frac{2n-\left(2n-2\right)}{1+2n\left(2n-2\right)}$

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

$\therefore$   Sum of the series $={\tan }^{-1}2-{\tan }^{-1}0+{\tan }^{-1}4-{\tan }^{-1}2+{\tan }^{-1}6-{\tan }^{-1}4+\dots {\tan }^{-1}2\mathrm{n}-{\tan }^{-1}\left(2\mathrm{n}-2\right)$

$={\tan }^{-1}2n-{\tan }^{-1}0={\tan }^{-1}2n$