Question

The sum of the series ${\cot }^{-1}(2^2+\frac{1}{2})+{\cot }^{-1}(2^3+\frac{1}{2^2})+{\cot }^{-1}(2^4+\frac{1}{2^3})+......\infty$ is equal to

• A. ${\tan }^{-1}\frac{1}{2}$
• B. ${\tan }^{-1}2$
• C. ${\tan }^{-1}2+{\tan }^{-1}3$
• D. ${\tan }^{-1}\frac{1}{2}+{\cot }^{-1}\frac{1}{2}$

Answer 1

The correct option is A ${\tan }^{-1}\frac{1}{2}$

We have,
${\cot }^{-1}(2^2+\frac{1}{2})+{\cot }^{-1}(2^3+\frac{1}{2^2})+{\cot }^{-1}(2^4+\frac{1}{2^3})+......\infty$
$T_n={\cot }^{-1}(2^{n+1}+\frac{1}{2^n})$
$={\cot }^{-1}\left(\frac{2^{n+1}\cdot 2^n+1}{2^n}\right)$
$={\tan }^{-1}\left(\frac{2^n(2-1)}{1+2^{n+1}\cdot 2^n}\right)$
$={\tan }^{-1}\left(\frac{2^{n+1}-2^n}{1+2^{n+1}\cdot 2^n}\right)$
Using the formula:
$\{{\tan }^{-1}\frac{x-y}{1+x\cdot y}={\tan }^{-1}x-{\tan }^{-1}y\}$
$={\tan }^{-1}{2}^{n+1}-{\tan }^{-1}{2}^n$
Putting $n=1,2,3,.....,n$ and adding, we get
$S_n={\tan }^{-1}{2}^{n+1}-{\tan }^{-1}2$
$\therefore S_{\infty }={\tan }^{-1}\infty -{\tan }^{-1}2$
$=\frac{\pi }{2}-{\tan }^{-1}2$
$={\cot }^{-1}2={\tan }^{-1}\frac{1}{2}$