已知正项等比数列an 的前n项和为sn ,bn=an^3/(an+1)^2,且bn的前n 项和Tn,若对一切

数列{an}为正项等比数列,则首项a1>0,公比q>0
bn=an³/a(n+1)²=[a1q^(n-1)]³/(a1qⁿ)²=a1q^(n-3)
b1=a1q^(1-3)=a1/q²
b(n+1)/bn=a1q^(n-2)/[a1q^(n-3)]=q,为定值,数列{bn}是以a1/q²为首项,q为公比的等比数列.
若q=1,则bn=a1 an=a1,Sn=Tn,与已知矛盾,因此q≠1
Sn>Tn
Sn/Tn>1
[a1(qⁿ-1)/(q-1)]/[(a1/q²)(qⁿ-1)/(q-1)]>1
q²>1
q>1或q1