Question

The value of ${\sum }_{m-1}^{n}ta{n}^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$ is,

• A. tan^-1(n²+n)
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is D

${\sum }_{m-1}^{n}ta{n}^{-1}\left(\frac{2m}{m^4+m^2+2}\right),={\sum }_{m-1}^{n}ta{n}^{-1}ta{n}^{-1}\left(\frac{2m}{1+(m^4+m^2+1)}\right)$
$={\sum }_{m-1}^{n}ta{n}^{-1}\left(\frac{2m}{1+(m^2+1{)}^2-m^2}\right)$
$={\sum }_{m-1}^{n}ta{n}^{-1}\left(\frac{2m}{1+(m^2+m+1).(m^2-m+1)}\right)$
$={\sum }_{m-1}^{n}ta{n}^{-1}\left(\frac{(m^2+m+1)-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}\right)$
$={\sum }_{m-1}^{n}ta{n}^{-1}\left\{ta{n}^{-1}\right(m^2+m+1)-ta{n}^{-1}(m^2-m+1\left)\right\}$
$=\left\{ta{n}^{-1}\right(3)-ta{n}^{-1}(1\left)\right\}+\left\{ta{n}^{-1}\right(7)-ta{n}^{-1}(3\left)\right\}+\cdots \cdots +\left\{ta{n}^{-1}\right(n^2+n+1)-ta{n}^{-1}(n^2-n-1)-ta{n}^{-1}(n^2-n-1\left)\right\}$
$=ta{n}^{-1}(n^2+n+1)-ta{n}^{-1}\left(1\right)$
$=ta{n}^{-1}\left(\frac{n^2+n+1-1}{1+(n^2+n+n+1)}\right)=ta{n}^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$