S = tan^{-1}left( frac{1}{n^2 + n + 1} right) | vedclass

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(A) We use the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.Each term in the series is of the form $	an^{-1}

Vedclass · Accepted Answer

$\frac{20}{n^2 + 20n + 1}$

Vedclass · Answer

(A) We use the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.Each term in the series is of the form $	an^{-1} \left( \frac{1}{1 + k(k+1)} ight) = 	an^{-1}(k+1) - 	an^{-1}(k)$.Let $f(k) = 	an^{-1}(k+1) - 	an^{-1}(k)$.The given sum is $S = \sum_{k=n}^{n+19} 	an^{-1} \left( \frac{1}{1 + k(k+1)} ight)$.$S = \sum_{k=n}^{n+19} (	an^{-1}(k+1) - 	an^{-1}(k))$.This is a telescoping sum:$S = (	an^{-1}(n+1) - 	an^{-1}(n)) + (	an^{-1}(n+2) - 	an^{-1}(n+1)) + \dots + (	an^{-1}(n+20) - 	an^{-1}(n+19))$.$S = 	an^{-1}(n+20) - 	an^{-1}(n)$.Using the formula $	an S = 	an(	an^{-1}(n+20) - 	an^{-1}(n)) = \frac{(n+20) - n}{1 + (n+20)n}$.$	an S = \frac{20}{1 + n^2 + 20n} = \frac{20}{n^2 + 20n + 1}$.

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$,then $\tan S$ is equal to

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