mathop {lim }limits_{n to infty } ,frac{{suml | vedclass

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(D) We know that $	an^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,so $	an^{-1}(1+r+r^2) = \frac{\pi}{2} - \cot^{-1}(1+r+r^2)$.Using the identity $\cot^{

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$\frac{\pi}{2}$

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(D) We know that $	an^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,so $	an^{-1}(1+r+r^2) = \frac{\pi}{2} - \cot^{-1}(1+r+r^2)$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$,we have $	an^{-1}(1+r+r^2) = \frac{\pi}{2} - 	an^{-1}(\frac{1}{1+r+r^2})$.Note that $\frac{1}{1+r+r^2} = \frac{(r+1)-r}{1+r(r+1)}$.Thus,$	an^{-1}(1+r+r^2) = \frac{\pi}{2} - (	an^{-1}(r+1) - 	an^{-1}(r))$.Summing from $r=0$ to $n$:$\sum_{r=0}^n 	an^{-1}(1+r+r^2) = \sum_{r=0}^n \frac{\pi}{2} - \sum_{r=0}^n (	an^{-1}(r+1) - 	an^{-1}(r))$.This is a telescoping sum: $\sum_{r=0}^n (	an^{-1}(r+1) - 	an^{-1}(r)) = 	an^{-1}(n+1) - 	an^{-1}(0) = 	an^{-1}(n+1)$.So,the expression becomes $\frac{(n+1)\pi}{2} - 	an^{-1}(n+1)$.Dividing by $n$ and taking the limit as $n 	o \infty$:$\lim_{n 	o \infty} \frac{\frac{(n+1)\pi}{2} - 	an^{-1}(n+1)}{n} = \lim_{n 	o \infty} (\frac{(n+1)\pi}{2n} - \frac{	an^{-1}(n+1)}{n}) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

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