The sum sumlimits_{n = 1}^infty {{cot }^{ - | vedclass

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(C) We know that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.Substituting this into the expression,we get:$\cot^{-1} \left( \frac{2 \cdot \frac{n(n+1)}{2} - 1

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$\pi$

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(C) We know that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.Substituting this into the expression,we get:$\cot^{-1} \left( \frac{2 \cdot \frac{n(n+1)}{2} - 1}{3} ight) = \cot^{-1} \left( \frac{n^2+n-1}{3} ight) = 	an^{-1} \left( \frac{3}{n^2+n-1} ight)$.We can rewrite the argument as $\frac{(n+2)-(n-1)}{1+(n+2)(n-1)} = \frac{3}{1+(n^2+n-2)}$.Thus,$T_n = 	an^{-1}(n+2) - 	an^{-1}(n-1)$.The sum is $S_N = \sum_{n=1}^N (	an^{-1}(n+2) - 	an^{-1}(n-1))$.Expanding the sum: $S_N = (	an^{-1} 3 - 	an^{-1} 0) + (	an^{-1} 4 - 	an^{-1} 1) + (	an^{-1} 5 - 	an^{-1} 2) + \dots + (	an^{-1}(N+2) - 	an^{-1}(N-1))$.Most terms cancel out,leaving $S_N = 	an^{-1}(N+2) + 	an^{-1}(N+1) - 	an^{-1} 1 - 	an^{-1} 0$.As $N 	o \infty$,$S = \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{4} = \frac{3\pi}{4}$.Since $\frac{3\pi}{4} = \frac{\pi}{2} + \frac{\pi}{4} = \frac{\pi}{2} + \cot^{-1} 1$,or using the options,$\frac{3\pi}{4} = \frac{\pi}{2} + 	an^{-1} 1$. Note that $	an^{-1} 1 = \cot^{-1} 1$. Checking the options,the sum is $\frac{3\pi}{4}$. Since $\cot^{-1} 2 + 	an^{-1} 2 = \frac{\pi}{2}$,we have $\frac{3\pi}{4} = \frac{\pi}{4} + \frac{\pi}{2} = \cot^{-1} 1 + \frac{\pi}{2}$. Option $A$ is $\frac{3\pi}{4} + \cot^{-1} 2$,which is not correct. Re-evaluating the sum: $S = \lim_{N 	o \infty} (	an^{-1}(N+2) + 	an^{-1}(N+1) - \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{4} = \frac{3\pi}{4}$. The correct value is $\frac{3\pi}{4}$.

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

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