Question

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

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

Answer 1

Answer: (B)

$ta{n}^{-1}\left(2^n\right)-\frac{\pi }{4}$

$\sum _{r=1}^n{\tan }^{-1}\left(\frac{2^{r-1}}{1+2^{2r-1}}\right)=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2^r-2^{r-1}}{1+2^r{.2}^{r-1}}\right)$
$\sum _{r=1}^n({\tan }^{-1}{2}^r-{\tan }^{-1}{2}^{r-1})=({\tan }^2-{\tan }^1)+({\tan }^{-1}3-{\tan }^{-1}2)+.......({\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1})$
$={\tan }^{-1}{2}^n-{\tan }^{-1}1={\tan }^{-1}{2}^n-\frac{\pi }{4}$