cot ^{-1}left(2 cdot 1^2right)+cot ^{-1}left( | vedclass

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(C) The general term of the series is $T_n = \cot^{-1}(2n^2)$.We know that $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$.So,$T_n = 	an^{-1}(\fr

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$\frac{\pi}{4}$

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(C) The general term of the series is $T_n = \cot^{-1}(2n^2)$.We know that $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$.So,$T_n = 	an^{-1}(\frac{1}{2n^2})$.We can rewrite the argument as $\frac{1}{2n^2} = \frac{2}{4n^2} = \frac{2}{1 + (2n^2 - 1)} = \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$.Using the formula $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}(\frac{x-y}{1+xy})$,we get:$T_n = 	an^{-1}(2n+1) - 	an^{-1}(2n-1)$.The sum of the first $N$ terms is $S_N = \sum_{n=1}^{N} [	an^{-1}(2n+1) - 	an^{-1}(2n-1)]$.This is a telescoping series:$S_N = (	an^{-1}(3) - 	an^{-1}(1)) + (	an^{-1}(5) - 	an^{-1}(3)) + \ldots + (	an^{-1}(2N+1) - 	an^{-1}(2N-1))$.$S_N = 	an^{-1}(2N+1) - 	an^{-1}(1)$.As $N 	o \infty$,$S_N = 	an^{-1}(\infty) - 	an^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

$\cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty =$

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