已知\(\left\{ {{a}_{n}} \right\}\)是等差数列,\(\left\{ {{b}_{n}} \right\}\)是等比数列,其中\({{a}_{1}}={{b}_{1}}=1\),\({{a}_{2}}+{{b}_{3}}={{a}_{4}}\),\({{a}_{3}}+{{b}_{4}}={{a}_{7}}\).
\((\)Ⅰ\()\)求数列\(\left\{ {{a}_{n}} \right\}\)与\(\left\{ {{b}_{n}} \right\}\)的通项公式;
\((\)Ⅱ\()\)记\({{c}_{n}}=\dfrac{1}{n}\left( {{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n}} \right)\left( {{b}_{1}}+{{b}_{2}}+\cdots +{{b}_{n}} \right)\),求数列\(\left\{ {{c}_{n}} \right\}\)的前\(n\)项和\({{S}_{n}}\).