已知数列\(\{a _{n} \}\)满足：\(a _{1} + \dfrac {1}{2}a_{2}+ \dfrac {1}{3}a_{3}+…+ \dfrac {1}{n}a_{n}= \dfrac {n^{2}+n}{2}(n∈N^{*})\)，数列\(\{b _{n} \}\)的前\(n\)项的和为\(S _{n}\)，\(S _{n} =2b _{n} -2\)，\(n∈N*\)．
\((1)\)求数列\(\{a _{n} \}\)和\(\{b _{n} \}\)的通项公式；
\((2)\)求数列\(\{a _{n} \boldsymbol{⋅}b _{n} \}\)的前\(n\)项和；
\((3)\)记\(c _{k} =b _{k} \boldsymbol{⋅}(b _{k} +b _{k+1} +…+b _{n} )(1\leqslant k\leqslant n)\)，数列\(\{c _{k} \}(1\leqslant k\leqslant n)\)的前\(n\)项的和为\(T _{n}\)，求证：\( \dfrac {2}{T_{1}} + \dfrac {2^{2}}{T_{2}} +…+ \dfrac {2^{n}}{T_{n}} < \dfrac {3}{4}\)．