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

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(A) The given series is $S = \sum_{r=1}^{\infty} \cot ^{-1}(2r^2) = \sum_{r=1}^{\infty} 	an ^{-1}\left(\frac{1}{2r^2}ight)$.We can rewrite the argu

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$\frac{\pi}{4}$

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(A) The given series is $S = \sum_{r=1}^{\infty} \cot ^{-1}(2r^2) = \sum_{r=1}^{\infty} 	an ^{-1}\left(\frac{1}{2r^2}ight)$.We can rewrite the argument of $	an^{-1}$ as:$\frac{1}{2r^2} = \frac{2}{4r^2} = \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$,we get:$S = \sum_{r=1}^{\infty} [	an^{-1}(2r+1) - 	an^{-1}(2r-1)]$.Expanding the summation:$S = (	an^{-1} 3 - 	an^{-1} 1) + (	an^{-1} 5 - 	an^{-1} 3) + (	an^{-1} 7 - 	an^{-1} 5) + \dots$This is a telescoping series where all intermediate terms cancel out.$S = \lim_{n 	o \infty} [	an^{-1}(2n+1) - 	an^{-1}(1)]$.Since $\lim_{n 	o \infty} 	an^{-1}(2n+1) = \frac{\pi}{2}$ and $	an^{-1}(1) = \frac{\pi}{4}$,$S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots$ up to $\infty$ is equal to

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