$\dfrac{1}{4.1}+\dfrac{1}{4.7}+\dfrac{1}{10.7}+... + \dfrac{1}{(3n-2)(3n+1)} \\ = \dfrac{1}{3}.[\dfrac{3}{4.1}+\dfrac{3}{4.7}+\dfrac{3}{10.7}+... + \dfrac{3}{(3n-2)(3n+1)}] \\ = \dfrac{1}{3}.(1- \dfrac{1}{4}+\dfrac{1}{4} - \dfrac{1}{7} ... + \dfrac{1}{3n-2} - \dfrac{1}{3n+1}) \\ = 1 - \dfrac{1}{3n+1} = \dfrac{3n}{3n+1}$