Question

${\sum }_{k=1}^{\infty }Ta{n}^{-1}\left(\frac{1}{2K^2}\right)=\theta ,thenTan\theta$ =___

• A. \N
• B. 1
• C. $\sqrt{3}$
• D. $\infty$

Answer 1

The correct option is B 1

$Ta{n}^{-1}\left[\frac{(2K+1)-(2K-1)}{1+(2K+1)(2K-1)}\right]=Ta{n}^{-1}(2K+1)-Ta{n}^{-1}(2K-1)$
Expanding $\sum$ we get, $Ta{n}^{-1}1$