已知等比数列{An}的公比q＜0,前n项和为Sn,则S4A5与S5A4的大小关系为

已知等比数列{An}的公比q＜0,前n项和为Sn,则S4A5与S5A4的大小关系为
已知等比数列{An}的公比q＜0,前n项和为Sn,则S4A5与S5A4的大小关系为

已知等比数列{An}的公比q＜0,前n项和为Sn,则S4A5与S5A4的大小关系为
S4=a1*(1-q^4)/(1-q)
A5=a1*q^4
S5=a1*(1-q^5)/(1-q)
A4=a1*q^3
所以有S4*A5-S5*A4=[a1*(1-q^4)*a1*q^4-a1*(1-q^5)*a1*q^3]/(1-q)
=a1^2*q^3*(q-q^5-1+q^5)/(1-q)
=-a1^2*q^3>0
故S4*A5>S5*A4