3.
设\(\{a_{n}\}\)是等比数列,公比大于\(0\),其前\(n\)项和为\(S_{n}(n∈N*)\),\(\{b_{n}\}\)是等差数列\(.\)已知\(a_{1}=1\),\(a_{3}=a_{2}+2\),\(a_{4}=b_{3}+b_{5}\),\(a_{5}=b_{4}+2b_{6}\).
\((\)Ⅰ\()\)求\(\{a_{n}\}\)和\(\{b_{n}\}\)的通项公式;
\((\)Ⅱ\()\)设数列\(\{S_{n}\}\)的前\(n\)项和为\(T_{n}(n∈N*)\),
\((i)\)求\(T_{n}\);
\((ii)\)证明\( \sum\limits_{k=1}^{n} \dfrac {(T_{k}+b_{k+2})b_{k}}{(k+1)(k+2)}= \dfrac {2^{n+2}}{n+2}-2(n∈N*)\).