已知等差数列\(\{a_{n}\}\)满足\(a_{1}=2\)，\(a_{4}+a_{16}=40.\)数列\(\{b_{n}\}\)的前\(n\)项和为\(S_{n}\)，且\(S_{n}=2^{n}-1.\)
\((1)\)求数列\(\{a_{n}\}\)与\(\{b_{n}\}\)的通项公式；
\((2)\)令\(c_{n}=a_{n}b_{n}\)，求数列\(\{c_{n}\}\)的前\(n\)项和\(T_{n}.\)