1)bn=3an+2
∴b﹤n-1﹥=3a﹤n-1﹥+2
bn-b﹤n-1﹥=3an+2-(3a﹤n-1﹥+2)=3(an-a﹤n-1﹥)=3*1=3
∴{bn}是以b1=1为首项,d=3为公差的等差数列
2)bn=b1+(n-1)d=1+(n-1)*3=3n-2