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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 24 of 160

Marks: +1, -0

cot

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

=

$\frac{10}{3}$
$\frac{8}{3}$
$\frac{14}{3}$
$\frac{16}{3}$

Solution:

Consider the expression,

\sum\limits_{k=1}^{n} 2k = n(n+1)

Then

\cot\left(\sum\limits_{n=3}^{32} \cot^{-1}(1+n(n+1))\right) = \sum\limits_{n=3}^{32} \tan^{-1}\left(\;\; \frac{(n+1)-n}{1+(n+1)n}\right)

This implies

\sum\limits_{n=3}^{32} \tan^{-1}\left(\;\; \frac{(n+1)-n}{1+(n+1)n}\right) = \sum\limits_{n=3}^{32} \tan^{-1}(n+1) - \tan^{-1} n

= (\tan^{-1}4 - \tan^{-1}3 + (\tan^{-1}5 - \tan^{-1}4)) + \dots

\& + (\tan^{-1}33 - \tan^{-2}32)

= \tan^{-1}33 - \tan^{-1}3

= \tan^{-1}\left[\;\; \frac{33-3}{1+99}\right]

Further simplify the above

\sum\limits_{n=3}^{32} \tan^{-1}\left(\;\; \frac{(n+1)-n}{1+(n+1)n}\right) = \tan^{-1}\left(\;\; \frac{3}{10}\right)

\cot\left(\sum\limits_{n=3}^{32} \tan^{-1}\left(\;\; \frac{(n+1)-n}{1+(n+1)n}\right)\right) = \cot\left(\tan^{-1}\left(\;\; \frac{3}{10}\right)\right)

\cot\left[\cot^{-1}\left(\;\; \frac{10}{3}\right)\right] = \;\; \frac{10}{3}

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