已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)满足：\(S_{n}=2(a_{n}-1)\)，数列\(\{b_{n}\}\)满足：对任意\(n∈N*\)有\(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}=(n-1)⋅2^{n+1}+2\)．
\((1)\)求数列\(\{a_{n}\}\)与数列\(\{b_{n}\}\)的通项公式；
\((2)\)记\(c_{n}= \dfrac {b_{n}}{a_{n}}\)，数列\(\{c_{n}\}\)的前\(n\)项和为\(T_{n}\)，证明：当\(n\geqslant 6\)时，\(n|T_{n}-2| < 1\)．