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The sum of the infinite terms of the series \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\tan ^{-1}\left(\frac{4}{33}\right)+\ldots is equal to

Option: 1

\frac{\pi}{6}


Option: 2

\frac{\pi}{4}


Option: 3

\frac{\pi}{3}


Option: 4

\frac{\pi}{2}


Answers (1)

best_answer

\quad \begin{aligned} \therefore \: \: \: \: T_r & =\tan ^{-1}\left(\frac{2^{r-1}}{1+2^{2 r-1}}\right) \\ \\& =\tan ^{-1}\left(\frac{2^r-2^{r-1}}{1+2^r \cdot 2^{r-1}}\right) \\ \\& =\tan ^{-1}\left(2^r\right)-\tan ^{-1}\left(2^{r-1}\right) \end{aligned}

\begin{aligned} \therefore S_n=\sum_{r=1}^n T_r & =\sum_{r=1}^n \tan ^{-1}\left(2^r\right)-\tan ^{-1}\left(2^{r-1}\right) \\ \\& =\tan ^{-1}\left(2^n\right)-\tan ^{-1}\left(2^0\right) \\ \\& =\tan ^{-1}\left(2^n\right)-\tan ^{-1}(1) \\ \\& =\tan ^{-1}\left(2^n\right)-\frac{\pi}{4} \end{aligned}

Hence,

                       \begin{aligned} S_{\infty} & =\tan ^{-1}\left(2^{\infty}\right)-\frac{\pi}{4} \\ \\& =\tan ^{-1}(\infty)-\frac{\pi}{4} \\ \\& =\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4} \end{aligned}

Posted by

Rishabh

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