Question

${\sum }_{m-1}^{\pi }ta{n}^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$ is equal to

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

We have ${\sum }_{m-1}^{\pi }ta{n}^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$
$={\sum }_{m-1}^{\pi }ta{n}^{-1}\left(\frac{2m}{1+(m^2+m+1)(m^2-m+1)}\right)={\sum }_{m-1}^{\pi }ta{n}^{-1}\left(\frac{(m^2+m+1)-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}\right)={\sum }_{m-1}^{\pi }\left[ta{n}^{-1}\right(m^2+m+1)-ta{n}^{-1}(m^2-m+1\left)\right]=(ta{n}^{-1}3-ta{n}^{-1})+(ta{n}^{-1}7-ta{n}^{-1}3)+(ta{n}^{-1}13-ta{n}^{-1}7)+.....+\left[ta{n}^{-1}\right(n^2+n+1\left)\right]\left(ta{n}^{-1}\right(n^2+n+1)-ta{n}^{-1}\left(\frac{n^2+n}{2+n^2+n}\right)$