Question

The value of the expression $co{t}^{-1}\frac{1}{2}+co{t}^{-1}\frac{9}{2}+co{t}^{-1}\frac{25}{2}+co{t}^{-1}\frac{49}{2}+......$ upto $n$ terms is

• A. $ta{n}^{-1}2n$
• B. ${\tan }^{-1}(2n-1)$
• C. $ta{n}^{-1}n$
• D. $ta{n}^{-1}2n-ta{n}^{-1}1$

Answer 1

Answer: (A)

Given expression
$=ta{n}^{-1}2+ta{n}^{-1}\frac{2}{9}+ta{n}^{-1}\frac{2}{25}+ta{n}^{-1}\frac{2}{49}+....$
General term
$=\frac{2}{{\left(2n-1\right)}^2}=\frac{2}{4n^2-4n+1}=\frac{2}{1+4n\left(n-1\right)}=\frac{2n-\left(2n-2\right)}{1+2n\left(2n-2\right)}$
$T_n={\tan }^{-1}2n-{\tan }^{-1}(2n-2)$
$\therefore$ Sum of the series
$={\tan }^{-1}2-{\tan }^{-1}0+{\tan }^{-1}4-{\tan }^{-1}2+{\tan }^{-1}6-{\tan }^{-1}4+\dots {\tan }^{-1}2n-{\tan }^{-1}(2n-2)$
$={\tan }^{-1}2n-{\tan }^{-1}0={\tan }^{-1}2n$