sum_{i=0}^2 cot ^{-1}{-(i+1)}= . . . . . . | vedclass

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(D) We need to evaluate the sum $S = \sum_{i=0}^2 \cot^{-1}(-(i+1))$.Expanding the sum,we get:$S = \cot^{-1}(-1) + \cot^{-1}(-2) + \cot^{-1}(-3)$.Usin

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$\frac{5 \pi}{2}$

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(D) We need to evaluate the sum $S = \sum_{i=0}^2 \cot^{-1}(-(i+1))$.Expanding the sum,we get:$S = \cot^{-1}(-1) + \cot^{-1}(-2) + \cot^{-1}(-3)$.Using the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$:$S = (\pi - \cot^{-1}(1)) + (\pi - \cot^{-1}(2)) + (\pi - \cot^{-1}(3))$$S = 3\pi - (\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3))$.We know $\cot^{-1}(1) = \frac{\pi}{4}$.For $\cot^{-1}(2) + \cot^{-1}(3)$,we use $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}(\frac{x+y}{1-xy})$ where $xy $\cot^{-1}(2) + \cot^{-1}(3) = 	an^{-1}(\frac{1}{2}) + 	an^{-1}(\frac{1}{3}) = 	an^{-1}(\frac{1/2 + 1/3}{1 - 1/6}) = 	an^{-1}(\frac{5/6}{5/6}) = 	an^{-1}(1) = \frac{\pi}{4}$.Thus,$S = 3\pi - (\frac{\pi}{4} + \frac{\pi}{4}) = 3\pi - \frac{\pi}{2} = \frac{5\pi}{2}$.

$\sum_{i=0}^2 \cot ^{-1}\{-(i+1)\}=$ . . . . . . .

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