The value of cot left( {sumlimits_{n = 1}^{19 | vedclass

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Question

(A) We know that $\sum_{p=1}^n 2p = 2 	imes \frac{n(n+1)}{2} = n^2 + n$. So,the expression inside the summation is $\cot^{-1}(1 + n^2 + n) = 	an^{-1

Vedclass · Accepted Answer

$\frac{21}{19}$

Vedclass · Answer

(A) We know that $\sum_{p=1}^n 2p = 2 	imes \frac{n(n+1)}{2} = n^2 + n$. So,the expression inside the summation is $\cot^{-1}(1 + n^2 + n) = 	an^{-1}\left(\frac{1}{1 + n(n+1)}ight)$. Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$,we can write: $	an^{-1}\left(\frac{(n+1) - n}{1 + n(n+1)}ight) = 	an^{-1}(n+1) - 	an^{-1}(n)$. Now,the sum becomes: $\sum_{n=1}^{19} (	an^{-1}(n+1) - 	an^{-1}(n)) = (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1} 20 - 	an^{-1} 19)$. This is a telescoping sum,which simplifies to $	an^{-1} 20 - 	an^{-1} 1$. Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$: $	an^{-1} 20 - 	an^{-1} 1 = 	an^{-1}\left(\frac{20-1}{1+20 	imes 1}ight) = 	an^{-1}\left(\frac{19}{21}ight)$. Finally,we need to find $\cot(	an^{-1}(19/21)) = \cot(\cot^{-1}(21/19)) = \frac{21}{19}$.

The value of $\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} } \right)$ is

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