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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

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Question : 24 of 160

Marks: +1, -0

if

\sum\limits_{k=1}^{n} \tan^{-1}

\left(\frac{1}{k^{2}+k+1}\right)

\tan^{-1}

(θ) , then θ =

$\frac{n}{n+2}$
$\frac{n}{n+1}$
1
$\frac{n}{n-1}$

Solution:

It is given that,

\sum\limits_{k=1}^{n} \tan^{-1}\left(\; \frac{1}{k^{2}+k+1}\right)=\tan^{-1}\theta

Simplifying the above expression, we get

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

\sum\limits_{k=1}^{n}\{\tan^{-1}(k+1)-\tan^{-1}k\}=\tan^{-1}\theta

\left\{\begin{array}{l}\tan^{-1}2-\tan^{-1}1+\tan^{-1}3-\tan^{-1}2+\tan^{-1}4- \\ \tan^{-1}3+\dots+\tan^{-1}(n+1)-\tan^{-1}n\end{array}\right\}=\tan^{-1}\theta

\tan^{-1}(n+1)-\tan^{-1}1=\tan^{-1}\theta

\tan^{-1}\left(\; \frac{n+1-1}{1+n+1}\right)=\tan^{-1}\theta

\frac{n}{2+n}=\theta

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