Question

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

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

Answer 1

Answer: (A)

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