A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{2\pi }{3}\]
D) None
Correct Answer: B
Solution :
[b] \[\frac{1}{1+r+{{r}^{2}}}=\frac{1}{1+r(r+1)}=\frac{\frac{1}{r(r+1)}}{1+\frac{1}{r(r+1)}}=\frac{\frac{1}{r}-\frac{1}{r+1}}{1+\frac{1}{r}\left( \frac{1}{r+1} \right)}\]\[\therefore {{\tan }^{-1}}\left( \frac{1}{1+r+{{r}^{2}}} \right)={{\tan }^{-1}}\frac{1}{r}-{{\tan }^{-1}}\frac{1}{r+1}\] \[\therefore \sum\limits_{r=1}^{\infty }{{{\tan }^{-1}}\left( \frac{1}{1+r+{{r}^{2}}} \right)}={{\tan }^{-1}}1=\frac{\pi }{4}\]
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