Question

Find the sum of the following series:
${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+....+{\tan }^{-1}\frac{2^{n-1}}{1+2^{2n-1}}$

• A. ${\tan }^{-1}{2}^n-{\tan }^{-1}1$
• B. $0$
• C. $\frac{\pi }{4}$
• D. ${\tan }^{-1}{2}^n-\frac{\pi }{4}$

Answer 1

Answer: (A), (D)

The correct options are
A ${\tan }^{-1}{2}^n-{\tan }^{-1}1$
D ${\tan }^{-1}{2}^n-\frac{\pi }{4}$

${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+.....+{\tan }^{-1}\frac{2^{n-1}}{1+2^{2n-1}}=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2^{r-1}}{1+2^{2r-1}}\right)=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2^r-2^{r-1}}{1+2^r.2^{r-1}}\right)\left(\because 2^r-2^{r-1}=2^{r-1}\right]\mathrm{We}\mathrm{know}\mathrm{that}{\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+x.y}\right)\mathrm{Therefore},{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+.....+{\tan }^{-1}\frac{2^{n-1}}{1+2^{2n-1}}=\sum _{r=1}^n\left({\tan }^{-1}{2}^r-{\tan }^{-1}{2}^{r-1}\right)=\left({\tan }^{-1}2-{\tan }^{-1}1\right)+\left({\tan }^{-1}{2}^2-{\tan }^{-1}2\right)+...+\left({\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1}\right)={\tan }^{-1}{2}^n-{\tan }^{-1}1={\tan }^{-1}{2}^n-\frac{\pi }{4}$