已知数列\(\{a_{n}\}\)，\(\{b_{n}\}\)，\(S_{n}\)为数列\(\{a_{n}\}\)的前\(n\)项和，\(a_{2}=4b_{1}\)，\(S_{n}=2a_{n}-2\)，\(nb_{n+1}-(n+1)b_{n}=n^{2}+n(n\in N^{*})\)
\((Ⅰ)\)求数列\(\{a_{n}\}\)的通项公式；
\((Ⅱ)\)证明\(\{\dfrac{b_{n}}{n}\}\)为等差数列；
\((Ⅲ)\)若数列\(\{c_{n}\}\)的通项公式为\(c_{n}=\begin{cases}-\dfrac{a_{n}b_{n}}{2},n\text{为奇数}\\ \dfrac {a_{n}b_{n}}{4},n\text{为偶数}\end{cases}\)，令\(T_{n}\)为\(\{c_{n}\}\)的前\(n\)项的和，求\(T_{2n}.\)