MCQ
$\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \frac{{{2}^{r-1}}}{1+{{2}^{2r-1}}} \right)=.......}$
- A${{\tan }^{-1}}{{2}^{n}}$
- ✓${{\tan }^{-1}}{{2}^{n}}-\frac{\pi }{4}$
- C${{\tan }^{-1}}{{2}^{n+1}}$
- D${{\tan }^{-1}}{{2}^{n+1}}-\frac{\pi }{4}$
$\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \frac{{{2}^{r-1}}}{1+{{2}^{2r-1}}} \right)}$
$= \sum _{r=1}^n \tan^{-1} \left(\frac {2^r - 2^{r-1}} {1+2^r 2^{r-1}}\right)$
$= \sum _{r=1}^n \left [ \tan^{-1} (2^r) - \tan^{-1} (2^{r-1}) \right]$
$= \tan^{-1} 2^n - \tan^{-1} 1$
$= \tan^{-1} 2^n - \frac{\pi}{4}$
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