\(=3x\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{870}\right)\)
=\(3x\left(\frac{1}{1x2}+\frac{1}{2x3}+....+\frac{1}{29x30}\right)\)
=\(3x\left(1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{30}\right)\)
\(=3x\left(1-\frac{1}{30}\right)=3x\frac{29}{30}=\frac{29}{10}\)