已知\(\left(1+x\right){2}^{n+1}={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+…+{a}_{2n+1}{x}^{2n+1} \)，\(n\in {{\mathbf{N}}^{*}}.\)记\({{T}_{n}}=\sum\limits_{k=0}^{n}{(\ 2k+1\ ){{a}_{n-k}}}\)．

\((1)\)求\(T_{2}\)的值；

\((2)\)化简\({{T}_{n}}\)的表达式，并证明：对任意的\(n\in {{\mathbf{N}}^{*}}\)，\({{T}_{n}}\)都能被\(4n+2\)整除．