Question

The value of ${\tan }^{-1}\left(\frac{1}{1+1+1^2}\right)+{\tan }^{-1}\left(\frac{1}{1+2+2^2}\right)+{\tan }^{-1}\left(\frac{1}{1+3+3^2}\right)+.......\infty$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{2\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{4}$

$Tn={\tan }^{-1}\left(\frac{1}{1+n+n^2}\right)={\tan }^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)$

$Tn={\tan }^{-1}(n+1)-{\tan }^{-1}n$

${\sum }_{n=1}^{\infty }Tn={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)$

$+{tan}^{-1}\left(3\right)-{tan}^{-1}\left(2\right)$

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${\sum }_{n=1}^{\infty }Tn={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(1\right)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$