设等比数列an的公比为q,前n项和Sn（n=1,2,3,……）1.求q的取值范围

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