已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且满足\(S_{n}=2a_{n}-2\)．
\((\)Ⅰ\()\)求数列\(\{a_{n}\}\)的通项公式；
\((\)Ⅱ\()\)设函数\(f(x)=( \dfrac {1}{2})^{x}\)，数列\(\{b_{n}\}\)满足条件\(b_{1}=2\)，\(f(b_{n+1})= \dfrac {1}{f(-3-b_{n})}\)，\((n∈N^{*})\)，若\(c_{n}= \dfrac {b_{n}}{a_{n}}\)，求数列\(\{c_{n}\}\)的前\(n\)项和\(T_{n}\)．

【考点】错位相减法,数列的递推关系,等比数列的通项公式

【分析】请登陆后查看

【解答】请登陆后查看