已知等比数列\(\{a _{n} \}\)的公比\(q > 1\)，且\(a _{2} +a _{3} +a _{4} =14\)，\(a _{3} +1\)是\(a _{2}\)，\(a _{4}\)的等差中项，数列\(\{b _{n} \}\)满足：数列\(\{a _{n} \boldsymbol{⋅}b _{n} \}\)的前\(n\)项和为\(n\boldsymbol{⋅}2 ^{n}\)．
\((1)\)求数列\(\{a _{n} \}\)、\(\{b _{n} \}\)的通项公式；
\((2)\)数列\(\{c _{n} \}\)满足：\(c _{1} =3\)，\(c _{n+1} =c _{n} + \dfrac {b_{n}}{c_{n}},n∈N^{*}\)，证明：\(c _{1} +c _{2} +…+c _{n} > \dfrac {n(n+2)}{2},n∈N^{*}\)．