12．已知数列{an}\{a_{n}\}{an}的前nnn项和Sn=3n−2(n{S}_{n}={3}^{n}-2(nSn=3n−2(n为正整数），则此数列的通项公式an=a_{n}=an=___{1,n=12⋅3n−1,n⩾2\left\{\begin{array}{l}{1,n=1}\\ {2\cdot {3}^{n-1},n\geqslant 2}\end{array}\right.{1,n=12⋅3n−1,n⩾2___．