设等比数列｛an｝公比为q,前n项和Sn>0（n＝1,2,3…）(1)求q的取值范围?

1)设等比数列通式an=a1q^(n-1)
显然a1大于零 【否则s10 所以00
2)bn=a(n+2)-(3/2)*a(n+1)=a(n)*q^2-(3/2)*a(n)*q
=a*[q^2-(3/2)*q]
所以Tn=Sn*[q^2-(3/2)*q
因为q>0
若q^2-(3/2)*q>1 即q>2时 Tn>Sn
若q^2-(3/2)*q=1 即q=2时 Tn=Sn
若q^2-(3/2)*q

【【【【解答】】】】（1）因为{an}是等比数列，Sn＞0,可得a1=S1＞0,q≠0.当q=1时,Sn=na1＞0. 当q≠1时,Sn=qqan−−1)1(1＞0，即qqn−−11＞0 (n=1,2,…),上式等价于不等式组 )1(01,01<−<−nqq或...