(Ⅰ)已知等比数列{an}中,设出公比为q,根据等比数列通项公式,代入a2=2,a5=128,求出公比,再利用等比数列的前n项和公式,代入Sn=35,求出n值;
(Ⅱ)因为bn=log2an,将an代入bn,求出其通项公式,代入不等式Sn<2bn求出n的范围;
【解析】
(Ⅰ)∵a2=a1q=2,a5=a1q4=128得q3=64,
∴q=4,a1=
∴an=a1qn-1==22n-3,∴bn=log2an=log222n-3=2n-3
∵bn+1-bn=[2(n+1)-3]-(2n-3)=2
∴{bn}是以b1=-1为首项,2为公差数列;
∴Sn==35,即n2-2n-35=0,可得(n-7)(n+5)=0,
即n=7;
(Ⅱ)∵Sn-bn=n2-2n-(2n-3)=n2-4n+3<0
∴3-<n<3+,∵n∈N+,
∴n=2,3,4,即所求不等式的解集为{2,3,4};