A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) None of these
Correct Answer: C
Solution :
[c] Let \[{{S}_{\infty }}={{\cot }^{-1}}2+{{\cot }^{-1}}8+{{\cot }^{-1}}18+{{\cot }^{-1}}32+...\] \[\therefore {{T}_{n}}={{\cot }^{-1}}2{{n}^{2}}={{\tan }^{-1}}\frac{1}{2{{n}^{2}}}\] \[={{\tan }^{-1}}\left( \frac{2}{4{{n}^{2}}} \right)={{\tan }^{-1}}\left( \frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)} \right)\] \[={{\tan }^{-1}}(2n+1)-ta{{n}^{-1}}(2n-1)\] \[\therefore \,\,\,\,\,{{S}_{n}}=\sum\limits_{n=1}^{\infty }{\{ta{{n}^{-1}}(2n+1)-{{\tan }^{-1}}(2n-1)\}}\] \[={{\tan }^{-1}}\infty -{{\tan }^{-1}}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\]
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