Question

$\sum _{r=1}^{10}\frac{r}{1-3r^2+r^4}$=

• A. $-\frac{25}{109}$
• B. $-\frac{35}{109}$
• C. $-\frac{45}{109}$
• D. $-\frac{55}{109}$

Answer 1

Answer: (D)

$-\frac{55}{109}$

$\frac{r}{1-3r^2+r^4}=\frac{r}{(r^2-1{)}^2-r^2}$
or $\frac{r}{(r^2-1-r)(r^2-1+r)}$
=$\frac{1}{2}[\frac{(r^2-1)+r-r^2-1-r}{(r^2-1-r)(r^2-1+r)}]$
$\therefore T_r=\frac{1}{2}[\frac{1}{r^2-1-r}-\frac{1}{r^2-1+r}]$
=$\frac{1}{2}[\frac{1}{1-r(r+1)}-\frac{1}{1-r(r-1)}]$
$\sum _{r=1}^{10}{T}_r=\frac{1}{2}[\frac{1}{1-10.11}-\frac{1}{1-1.0}]=-\frac{55}{109}$
The terms will cancel diagonally except the first and the last.