已知数列\(\{a _{n} \}\)，\(\{b _{n} \}\)满足\(2S _{n} =(a _{n} +2)b _{n}\)，其中\(S _{n}\)是数列\(\{a _{n} \}\)的前\(n\)项和．
\((1)\)若数列\(\{a _{n} \}\)是首项为\( \dfrac {2}{3}\)，公比为\( \dfrac {1}{3}\)的等比数列，求数列\(\{b _{n} \}\)的通项公式；
\((2)\)若\(b _{n} =n\)，\(a _{2} =3\)．
\((ⅰ)\)求\(\{a _{n} \}\)通项公式；
\((ⅱ)\)求证：\( \dfrac {1}{b_{1}a_{2}}+ \dfrac {1}{b_{2}a_{4}}+…+ \dfrac {1}{b_{n}a_{2n}} < 1\)．