(Ⅰ)设数列{an}的公差为d≠0,∵a1,a2,a6成等比数列,∴a22=a1a6,∴(1+d)2=1×(1+5d),化为d2-3d=0,∵d≠0,∴d=3,∴an=1+3(n-1)=3n-2.(2)∵等比数列{bn}的首项为1,公比q= a2 a1 =4,∴b1+b2+…+bk=1+4+…+4k-1= 1?4k 1?4 =85,化为4k=256,解得k=4.