If S is the sum of the first 10 terms of the | vedclass

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(D) The $n$-th term of the series is $T_n = 	an ^{-1}\left(\frac{1}{n^2+n+1}ight)$.We can rewrite this as $T_n = 	an ^{-1}\left(\frac{(n+1)-n}{1+n

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$\frac{5}{6}$

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(D) The $n$-th term of the series is $T_n = 	an ^{-1}\left(\frac{1}{n^2+n+1}ight)$.We can rewrite this as $T_n = 	an ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}ight) = 	an ^{-1}(n+1) - 	an ^{-1}(n)$.For the first $10$ terms,the sum $S$ is given by:$S = \sum_{n=1}^{10} (	an ^{-1}(n+1) - 	an ^{-1}(n))$.This is a telescoping series:$S = (	an ^{-1}(2) - 	an ^{-1}(1)) + (	an ^{-1}(3) - 	an ^{-1}(2)) + \ldots + (	an ^{-1}(11) - 	an ^{-1}(10))$.$S = 	an ^{-1}(11) - 	an ^{-1}(1)$.Using the formula $	an ^{-1}(x) - 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x-y}{1+xy}ight)$:$S = 	an ^{-1}\left(\frac{11-1}{1+11 	imes 1}ight) = 	an ^{-1}\left(\frac{10}{12}ight) = 	an ^{-1}\left(\frac{5}{6}ight)$.Therefore,$	an (S) = \frac{5}{6}$.

If $S$ is the sum of the first $10$ terms of the series $\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\tan ^{-1}\left(\frac{1}{21}\right)+\ldots$ then $\tan ( S )$ is equal to

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