cot left[sum_{n=3}^{32} cot ^{-1}left(1+sum_{ | vedclass

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(A) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Substituting this into the expression,we get $\cot \left[\sum_{n=3}^{32} \cot

Vedclass · Accepted Answer

$\frac{10}{3}$

Vedclass · Answer

(A) We know that $\sum_{k=1}^n 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Substituting this into the expression,we get $\cot \left[\sum_{n=3}^{32} \cot ^{-1}(1+n(n+1))ight]$.Using the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$,we have $\cot^{-1}(1+n(n+1)) = 	an^{-1}\left(\frac{1}{1+n(n+1)}ight)$.Since $\frac{1}{1+n(n+1)} = \frac{(n+1)-n}{1+(n+1)n}$,we can write this as $	an^{-1}(n+1) - 	an^{-1}(n)$.Thus,the sum becomes $\sum_{n=3}^{32} [	an^{-1}(n+1) - 	an^{-1}(n)]$.This is a telescoping sum: $(	an^{-1} 4 - 	an^{-1} 3) + (	an^{-1} 5 - 	an^{-1} 4) + \dots + (	an^{-1} 33 - 	an^{-1} 32)$.After cancellation,we are left with $	an^{-1} 33 - 	an^{-1} 3$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$,we get $	an^{-1}\left(\frac{33-3}{1+33 	imes 3}ight) = 	an^{-1}\left(\frac{30}{100}ight) = 	an^{-1}\left(\frac{3}{10}ight)$.Finally,we need to evaluate $\cot(	an^{-1}(\frac{3}{10}))$.Since $\cot(	an^{-1}(x)) = \cot(\cot^{-1}(\frac{1}{x})) = \frac{1}{x}$,we have $\cot(	an^{-1}(\frac{3}{10})) = \frac{10}{3}$.Therefore,the correct option is $A$.

$\cot \left[\sum_{n=3}^{32} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right]=$

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