\(=\dfrac{1}{3}-\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{33\cdot35}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{33}-\dfrac{1}{35}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{32}{105}=\dfrac{19}{105}\)