(1)设{an}的公差为d,数列{bn}的公比为q,由于a1=b1=1且a2=b1+1,a3=b3+1,则 1+d=1+q 1+2d=1+q2 ,解得d=q=2,则an=2n-1,bn=2n-1;(2)Sn=1+2+22+…+2n-1= 1?2n 1?2 =2n-1,则Sn- an+1 n =2n-1- 2n?1+1 n =2n-3>100∴2n>103,∵n是正整数∴满足要求的最小正整数n是7.