Practicing Success
$\sum\limits_{r=1}^{n}\tan^{-1}(\frac{2^{r-1}}{1+2^{2r-1}})$ is equal to: |
$\tan^{-1}(2^n)$ $\tan^{-1}(2^n)-\frac{π}{4}$ $\tan^{-1}(2^{n+1})$ $\tan^{-1}(2^{n+1})-\frac{π}{4}$ |
$\tan^{-1}(2^n)-\frac{π}{4}$ |
$\sum\limits_{r=1}^{n}\tan^{-1}(\frac{2^{r-1}}{1+2^{2r-1}})=\sum\limits_{r=1}^{n}\tan^{-1}(\frac{2^r-2^{r-1}}{1+2^r.2^{r-1}})=\sum\limits_{r=1}^{n}\tan^{-1}2^r-\sum\limits_{r=1}^{n}\tan^{-1}2^{r-1}=\tan^{-1}2^n-\tan^{-1}1=\tan^{-1}2^n-\frac{π}{4}$ |