If operatorname{Tan}^{-1} frac{1}{3}+operator | vedclass

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(A) The general term of the series is $T_k = \operatorname{Tan}^{-1} \frac{1}{k^2+k+1}$.We can rewrite the argument as $\frac{1}{1+k(k+1)}$.Using the

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$\frac{n}{n+2}$

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(A) The general term of the series is $T_k = \operatorname{Tan}^{-1} \frac{1}{k^2+k+1}$.We can rewrite the argument as $\frac{1}{1+k(k+1)}$.Using the identity $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \frac{x-y}{1+xy}$,we have:$T_k = \operatorname{Tan}^{-1} (k+1) - \operatorname{Tan}^{-1} k$.Summing from $k=1$ to $n$:$S_n = \sum_{k=1}^{n} (\operatorname{Tan}^{-1} (k+1) - \operatorname{Tan}^{-1} k) = \operatorname{Tan}^{-1} (n+1) - \operatorname{Tan}^{-1} (1)$.Using the formula $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \frac{x-y}{1+xy}$:$S_n = \operatorname{Tan}^{-1} \frac{(n+1)-1}{1+(n+1)(1)} = \operatorname{Tan}^{-1} \frac{n}{n+2}$.Thus,$	heta = \frac{n}{n+2}$.

If $\operatorname{Tan}^{-1} \frac{1}{3}+\operatorname{Tan}^{-1} \frac{1}{7}+\operatorname{Tan}^{-1} \frac{1}{13}+\ldots+\operatorname{Tan}^{-1} \frac{1}{n^2+n+1}=\operatorname{Tan}^{-1} \theta$,then $\theta=$

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