$VT=\frac{1}{3}(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{(3n-1)(3n+2)})$$=\frac{1}{3}.(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2})$
$=\frac{1}{3}(\frac{1}{2}-\frac{1}{3n+2})=\frac{1}{3}.\frac{3n}{6n+4}=\frac{n}{6n+4}$