Question

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

• A. ${\tan }^{-1}\left(2^n\right)$
• B. ${\tan }^{-1}\left(2^n\right)-\pi /4$
• C. ${\tan }^{-1}\left(2^{n+1}\right)$
• D. ${\tan }^{-1}\left(2^{n+1}\right)+\pi /4$

Answer 1

The correct option is B ${\tan }^{-1}\left(2^n\right)-\pi /4$

Given, $\sum _{r=1}^n{\tan }^{-1}\left(\frac{2^{r-1}}{1+2^{2r-1}}\right)$

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

$={\sum }_{r=1}^n[{\tan }^{-1}{2}^r-{\tan }^{-1}{2}^{r-1}]$
$=({\tan }^{-1}2-{\tan }^{-1}{2}^0)+({\tan }^{-1}{2}^2-{\tan }^{-1}{2}^1)+........+({\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1})$
$={\tan }^{-1}{2}^n-{\tan }^{-1}1$

$={\tan }^{-1}{2}^n-\frac{\pi }{4}$