The sum of the first n terms of the series co | vedclass

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(D) The $r$-th term of the series is $T_r = \cot^{-1}(r^2 + r + 1)$.Using the identity $\cot^{-1} x = 	an^{-1} \left( \frac{1}{x} ight)$,we have $T

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All of these

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(D) The $r$-th term of the series is $T_r = \cot^{-1}(r^2 + r + 1)$.Using the identity $\cot^{-1} x = 	an^{-1} \left( \frac{1}{x} ight)$,we have $T_r = 	an^{-1} \left( \frac{1}{r^2 + r + 1} ight)$.We can rewrite the denominator as $1 + r(r+1)$,so $T_r = 	an^{-1} \left( \frac{(r+1) - r}{1 + r(r+1)} ight)$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$,we get $T_r = 	an^{-1}(r+1) - 	an^{-1} r$.The sum of the first $n$ terms is $S_n = \sum_{r=1}^{n} (	an^{-1}(r+1) - 	an^{-1} r)$.This is a telescoping series: $S_n = (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1}(n+1) - 	an^{-1} n)$.$S_n = 	an^{-1}(n+1) - 	an^{-1} 1$.Using $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$,$S_n = 	an^{-1} \left( \frac{n+1-1}{1+(n+1)(1)} ight) = 	an^{-1} \left( \frac{n}{n+2} ight)$.Since $	an^{-1} x = \cot^{-1} (1/x)$,$S_n = \cot^{-1} \left( \frac{n+2}{n} ight)$.Thus,all options are equivalent.

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

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