Question

If $S=\sum _{r=1}^{80}\frac{r}{\left(r^4+r^2+1\right)},$ then the value of $\frac{6481S}{324}$ is

Answer 1

Answer: 1

$T_r=\frac{1}{2}\left\{\frac{\left(r^2+r+1\right)-\left(r^2-r+1\right)}{\left(r^2+r+1\right)\left(r^2-r+1\right)}\right\}$
$=\frac{1}{2}\left\{\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right\}$
$=\frac{1}{2}\left\{\frac{1}{r^2-r+1}-\frac{1}{(r+1{)}^2-(r+1)+1}\right\}$
$=\frac{1}{2}\{f(r)-f(r+1)\}$
$S=\Sigma T_r={\sum }_{r=1}^{80}\frac{1}{2}(f(\alpha )-f(\alpha +1))$
$=\frac{1}{2}\{f(1)-f(2)+f(2)-f(3)\dots +f(80)-f(81)\}$
$=\frac{1}{2}(f(1)-f(81))=\frac{1}{2}\left(1-\frac{1}{81^2-81+1}\right)$
$=\frac{1}{2}\left(1-\frac{1}{6481}\right)\Rightarrow 6481S=3240$
$=\frac{6481S}{3240}=1$