Question

Find the sum to $n$ terms
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....$

Answer 1

$\sum _{r=1}^n\frac{1}{(3r-1)(3r+2)}=\sum _{r=1}^n\frac{1}{3}\frac{(3r+2)-(3r-1)}{(3r-1)(3r+2)}$

$\frac{1}{3}\sum _{r=1}^n\frac{1}{3r-1}-\frac{1}{3r+2}=\frac{1}{3}(\frac{1}{3-1}-\frac{1}{3n+2})$

$=\frac{1}{3}(\frac{1}{2}-\frac{1}{3n+2})$

$=\frac{1}{6}-\frac{1}{9n+6}$