已知数列\(\{a _{n} \}\)的前\(n\)项和为\(S _{n}\)，已知\(a _{1} =2\)，\(6S _{n} =3na _{n+1} -2n(n+1)(n+2)\)，\(n∈N*\)．
\((1)\)求数列\(\{a _{n} \}\)的通项公式；
\((2)\)证明：\( \dfrac {1}{a_{1}}+ \dfrac {1}{a_{2}}+…+ \dfrac {1}{a_{n}} < \dfrac {5}{6}\)．