Question

$\displaystyle \sum _{r = 1}^{n} \frac{r}{r^{4} + r^{2} + 1}$ is equal to

• A. $\frac{n^{2} + n}{2 \left(n^{2} + n + 1\right)}$
• B. $\frac{n^{2} + 2 n}{2 \left(n^{2} + n + 1\right)}$
• C. $\frac{2 n^{2} + n}{2 \left(n^{2} + n + 1\right)}$
• D. $\frac{n^{2} + n}{\left(n^{2} + n + 1\right)}$

Answer 1

Answer: (A)

Let, $Ι\left(r\right)=\frac{r}{r^{4} + r^{2} + 1}=\frac{r}{\left(r^{2} + r + 1\right) \left(r^{2} - r + 1\right)}$
$=\frac{1}{2}\left\{\frac{\left(r^{2} + r + 1\right) - \left(r^{2} - r + 1\right)}{\left(r^{2} + r + 1\right) \left(r^{2} - r + 1\right)}\right\}$
$=-\frac{1}{2}\left(\frac{1}{r^{2} + r + 1} - \frac{1}{r^{2} - r + 1}\right)$
$\left.=-\frac{1}{2}(V r)-V(r-1)\right)$
$\Rightarrow \displaystyle\sum_{r=1}^{n} I (r)=-\frac{1}{2}(V(n)-V(0))$
$=-\frac{1}{2}\left(\frac{1}{n^{2}+n+1}-1\right)$
$=\frac{1}{2}\left(1-\frac{1}{n^{2}+n+1}\right)=\frac{1}{2} \frac{\left(n^{2}+n\right)}{\left(n^{2}+n+1\right)}$