Show that $$\dfrac{3}{1\cdot2\cdot4}+\dfrac{4}{2\cdot3\cdot5}+\dfrac{5}{3\cdot4\cdot6}+\cdots+\dfrac{n+2}{n\cdot(n+1)\cdot(n+3)}\\=\dfrac{1}{6}\left[\dfrac{29}{6}-\dfrac{4}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3} \right]$$
Trial: I proceed in this way $$\dfrac{3}{1\cdot2\cdot4}+\dfrac{4}{2\cdot3\cdot5}+\dfrac{5}{3\cdot4\cdot6}+\cdots+\dfrac{n+2}{n\cdot(n+1)\cdot(n+3)}\\=\dfrac{1}{6} \left[\dfrac{18}{1\cdot2\cdot4}+\dfrac{24}{2\cdot3\cdot5}+\dfrac{30}{3\cdot4\cdot6}+\cdots+\dfrac{6n+12}{n\cdot(n+1)\cdot(n+3)} \right] $$Then stuck how to break $6n+12$ so that we get the following result.