证明：(1)an=Sn-Sn-1=n^2+2n -(n-1)^2-2(n-1)=2n+1(n≥1)又S1=a1=3,an-an-1=2所以{an}是是以公差为2的等差数列,通项公式为an=2n+1(n≥1).(2)bn=2^(2n+1),所以bn-1=2^(2n-1)所以bn/bn-1=4所以b1=8,bn=8×4^(n-1) (n≥...