Question

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

Answer 1

$T_{r}=\frac{1}{2}\left\{\frac{\left(r^{2}+r+1\right)-\left(r^{2}-r+1\right)}{\left(r^{2}+r+1\right)\left(r^{2}-r+1\right)}\right\}$
$=\frac{1}{2}\left\{\frac{1}{r^{2}-r+1}-\frac{1}{r^{2}+r+1}\right\}$
$=\frac{1}{2}\left\{\frac{1}{r^{2}-r+1}-\frac{1}{(r+1)^{2}-(r+1)+1}\right\}$
$=\frac{1}{2}\{f(r)-f(r+1)\}$
$S=\Sigma T_{r}=\sum_{r=1}^{80} \frac{1}{2}(f(\alpha)-f(\alpha+1))$
$=\frac{1}{2}\{f(1)-f(2)+f(2)-f(3) \ldots+f(80)-f(81)\}$
$=\frac{1}{2}(f(1)-f(81))=\frac{1}{2}\left(1-\frac{1}{81^{2}-81+1}\right)$
$=\frac{1}{2}\left(1-\frac{1}{6481}\right) \Rightarrow 6481 S=3240$
$=\frac{6481 S}{3240}=1$