Tn=cot−1⁡n2+34=tan−1⁡44n2+3=tan−1⁡1n2+34=tan−1⁡11+n2−14=tan−1⁡n+12−n−121+n+12n−12=tan−1⁡n+12−tan−1⁡n−12∴cot−1⁡12+34+cot−1⁡22+34+cot−1⁡32+34+…….=∑n=1∞ Tn=∑n=1∞ tan−1⁡n+12−tan−1⁡n−12=tan−1⁡32−tan−1⁡12−tan−1⁡52−tan−1⁡32+tan−1⁡72−tan−1⁡52+………+tan−1⁡(∞)=tan−1⁡(∞)−tan−1⁡12=π2−tan−1⁡12=cot−1⁡12=tan−1⁡(2)

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Sum of the infinite terms of the series $\cot^{- 1} (1^{2} + \frac{3}{4}) + \cot^{- 1} (2^{2} + \frac{3}{4}) + \cot^{- 1} (3^{2} + \frac{3}{4}) + \dots \dots .$

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$\tan^{- 1} 3$

$\tan^{- 1} 2$

$\frac{π}{4}$

$\tan^{- 1} 4$

answer is B.

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Detailed Solution

$\begin{array}{l} T_{n} = \cot^{- 1} (n^{2} + \frac{3}{4}) = \tan^{- 1} (\frac{4}{4 n^{2} + 3}) = \tan^{- 1} (\frac{1}{n^{2} + \frac{3}{4}}) = \tan^{- 1} (\frac{1}{1 + n^{2} - \frac{1}{4}}) = \tan^{- 1} (\frac{(n + \frac{1}{2}) - (n - \frac{1}{2})}{1 + (n + \frac{1}{2}) (n - \frac{1}{2})}) = \tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} (n - \frac{1}{2}) \\ ∴ \cot^{- 1} (1^{2} + \frac{3}{4}) + \cot^{- 1} (2^{2} + \frac{3}{4}) + \cot^{- 1} (3^{2} + \frac{3}{4}) + \dots \dots . = \sum_{n = 1}^{\infty} T_{n} = \sum_{n = 1}^{\infty} (\tan^{- 1} (n + \frac{1}{2}) - \tan^{- 1} (n - \frac{1}{2})) \\ = \tan^{- 1} (\frac{3}{2}) - \tan^{- 1} (\frac{1}{2}) - \tan^{- 1} (\frac{5}{2}) - \tan^{- 1} (\frac{3}{2}) + \tan^{- 1} (\frac{7}{2}) - \tan^{- 1} (\frac{5}{2}) + \dots \dots \dots + \tan^{- 1} (\infty) \\ = \tan^{- 1} (\infty) - \tan^{- 1} (\frac{1}{2}) = \frac{π}{2} - \tan^{- 1} (\frac{1}{2}) = \cot^{- 1} (\frac{1}{2}) = \tan^{- 1} (2) \end{array}$