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Question Number 22940 by selestian last updated on 24/Oct/17

Commented by ajfour last updated on 24/Oct/17

$$T_r={\tan }^{-1}\left(\frac{2^{r-1}}{1+2^{2r-1}}\right)$$$$\Rightarrow \tan T_r=\frac{2^r-2^{r-1}}{1+2^r.2^{r-1}}$$$$=\tan ({\theta }_r-{\theta }_{r-1})$$$$suchthat\tan {\theta }_r=2^r$$$$So{T}_r={\theta }_r-{\theta }_{r-1}$$$$={\tan }^{-1}\left(2^r\right)-{\tan }^{-1}\left(2^{r-1}\right)$$$$\underset{r=1}{\sum ^n}{T}_r={\tan }^{-1}\left(2^n\right)-{\tan }^{-1}1$$$$={\tan }^{-1}\left(2^n\right)-\frac{\pi }{4}.$$

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