Question

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

• A. $ta{n}^{-1}(n^2+n+1)$
• B. $ta{n}^{-1}(n^2-n+1)$
• C. $ta{n}^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$
• D. None of these

Answer 1

The correct option is C $ta{n}^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$

$\sum _{m=1}^{n}ta{n}^{-1}\frac{2m}{1+(m^2+m+1)(m^2-m+1)}$
$\sum _{m=1}^{n}ta{n}^{-1}\frac{m^2+m+1-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}$
$=\sum _{m=1}^n\left(ta{n}^{-1}\right(m^2+m+1)-ta{n}^{-1}(m^2-m+1\left)\right)$
$=(ta{n}^{-1}3-ta{n}^{-1}1)+(ta{n}^{-1}7-ta{n}^{-1}3)+...+\left(ta{n}^{-1}\right(n^2+n+1)-ta{n}^{-1}(n^2-n+1\left)\right)$
We can observe that this forms a telescopic series since $(m^2+m+1=(m+1{)}^2-(m+1)+1)$
$S=ta{n}^{-1}(n^2+n+1)-ta{n}^{-1}\left(1\right)$
$S=ta{n}^{-1}\left(\frac{n^2+n}{n^2+n+2}\right)$