$T a n^{- 1} (\frac{2}{2 + 1^{2} + 1^{4}}) + T a n^{- 1} (\frac{4}{2 + 2^{2} + 2^{4}}) + T a n^{- 1} (\frac{6}{2 + 3^{2} + 3^{4}}) + ....... \infty$ terms

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$\frac{9 π}{2}$

$\frac{π}{4}$

$\frac{3 π}{2}$

$\frac{π}{2}$

answer is C.

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Detailed Solution

$T_{n} = T a n^{- 1} (\frac{2}{2 + n^{2} + n^{4}}) [T a n^{- 1} (x) - T a n^{- 1} (y) = T a n^{- 1} (\frac{x - y}{1 + x y})]$

$= T a n^{- 1} [\frac{2 n}{1 + (n^{4} + n^{2} + 1)}] = T a n^{- 1} [\frac{2 n}{1 + (n^{2} + n + 1) (n^{2} - n + 1)}] T_{n} = T a n^{- 1} (n^{2} + n + 1) - T a n^{- 1} (n^{2} - n + 1) T_{1} = T a n^{- 1} (3) - T a n^{- 1} (1) T_{2} = T a n^{- 1} (7) - T a n^{- 1} (3)$

………………………..

$T_{n} = T a n^{- 1} (n^{2} + n + 1) - T a n^{- 1} (n^{2} - n + 1) S_{n} = T a n^{- 1} (n^{2} + n + 1) - T a n^{- 1} (1) A S n \to \infty S_{\infty} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$