$\sum\limits_{r=1}^{n}\tan^{-1}(\frac{2^{r-1}}{1+2^{2r-1}})$ is equal to:
Answer & explanation
Correct answer: option 2
$\sum\limits_{r=1}^{n}\tan^{-1}(\frac{2^{r-1}}{1+2^{2r-1}})=\sum\limits_{r=1}^{n}\tan^{-1}(\frac{2^r-2^{r-1}}{1+2^r.2^{r-1}})=\sum\limits_{r=1}^{n}\tan^{-1}2^r-\sum\limits_{r=1}^{n}\tan^{-1}2^{r-1}=\tan^{-1}2^n-\tan^{-1}1=\tan^{-1}2^n-\frac{π}{4}$