If cot ^{-1}(alpha)=cot ^{-1} 2+cot ^{-1} 8+c | vedclass

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(A) The given series is $\sum_{n=1}^{100} \cot^{-1}(2n^2)$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$,we have $\sum_{n=1}^{100} 	an^{

Vedclass · Accepted Answer

$1.01$

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(A) The given series is $\sum_{n=1}^{100} \cot^{-1}(2n^2)$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$,we have $\sum_{n=1}^{100} 	an^{-1}(\frac{1}{2n^2})$.Multiplying numerator and denominator by $2$,we get $\sum_{n=1}^{100} 	an^{-1}(\frac{2}{4n^2})$.This can be written as $\sum_{n=1}^{100} 	an^{-1}(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)})$.Using the formula $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}(\frac{x-y}{1+xy})$,the sum becomes $\sum_{n=1}^{100} (	an^{-1}(2n+1) - 	an^{-1}(2n-1))$.This is a telescoping series: $(	an^{-1} 3 - 	an^{-1} 1) + (	an^{-1} 5 - 	an^{-1} 3) + \dots + (	an^{-1} 201 - 	an^{-1} 199)$.After cancellation,we are left with $	an^{-1} 201 - 	an^{-1} 1$.Using the formula again,$	an^{-1}(\frac{201-1}{1+201 	imes 1}) = 	an^{-1}(\frac{200}{202}) = 	an^{-1}(\frac{100}{101})$.Since $\cot^{-1}(\alpha) = 	an^{-1}(\frac{101}{100})$,we have $\alpha = \frac{101}{100} = 1.01$.

If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots$ up to $100$ terms,then $\alpha$ is

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