Question

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

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

Answer 1

The correct option is A ${tan}^{-1}\left(2^n\right)$

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

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

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

$={\text{tan}}^{-1}2-{\text{tan}}^{-1}1+{\text{tan}}^{-1}{2}^2-{\text{tan}}^{-1}{2}^1+\cdots +{\text{tan}}^{-1}{2}^n-{\text{tan}}^{-1}{2}^{n-1}$

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