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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 23 of 160

Marks: +1, -0

If

\sum\limits_{n=1}^{k} \tan^{-1}\left(\frac{1}{n^{2}+3n+3}\right)=\tan^{-1} \alpha

\alpha =

$\frac{k}{k+2}$
$\frac{2k}{2k+1}$
$\frac{k}{2k+5}$
$\frac{3k}{4k+5}$

Solution:

We have,

\sum\limits_{n=1}^{k} \tan^{-1}\left(\frac{1}{n^{2}+3n+3}\right)=\tan^{-1} \alpha

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

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

\Rightarrow \; \; (\tan^{-1}3-\tan^{-1}2)+(\tan^{-1}4-\tan^{-1}2)\dots

\tan^{-1}(k+2)-\tan^{-1}(k+1)=\tan^{-1} \alpha

\tan^{-1}(k+2)-\tan^{-1}2=\tan^{-1} \alpha

\tan^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right)=\tan^{-1} \alpha

\Rightarrow \; \; \; \; \frac{k}{1+2k+4}=\alpha \Rightarrow \alpha=\frac{k}{2k+5}

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