已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且\(\{\dfrac{S_{n}}{n}\}\)是等差数列，\(a_{1}=2\)，\(a_{2}=4.\)
\((1)\)求数列\(\{a_{n}\}\)的通项公式；
\((2)\)若数列\(\{b_{n}\}\)满足\(b_{n}=\dfrac{1}{(a_{n}-1)(2n+1)}\)，求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}.\)