6.
已知数列\(\left\{ {{a}_{n}} \right\}\)的前\(n\)和为\({{S}_{n}}\),且满足\({{S}_{2}}=3,\begin{matrix} {} \\ \end{matrix}2{{S}_{n}}=n+n{{a}_{n}},\begin{matrix} {} \\ \end{matrix}n\in {{N}^{*}}\)
\((1)\)求\({{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}\),并求出数列\(\left\{ {{a}_{n}} \right\}\)的通项公式;
\((2)\)设\({{b}_{n}}={{2}^{{{a}_{n}}+1}}-1\),证明:\(\dfrac{{{b}_{1}}}{{{b}_{2}}}+\dfrac{{{b}_{2}}}{{{b}_{3}}}+\cdots +\dfrac{{{b}_{n}}}{{{b}_{n+1}}} < \dfrac{n}{2}(n\in {{N}^{*}})\)