若等比数列{an}的前n项和为Sn,则Sn,S2n-Sn,S3n-S2n满足关系式（S2n-Sn)^2=Sn(S3n-S2n),但Sn,S2n-Sn,S3n-S2n不一定成等比数列.

Question

若等比数列{an}的前n项和为Sn,则Sn,S2n-Sn,S3n-S2n满足关系式
（S2n-Sn)^2=Sn(S3n-S2n),但Sn,S2n-Sn,S3n-S2n不一定成等比数列.

Answer 1

Sn=a1(1-q^n)/(1-q)
S2n=a1(1-q^2n)/(1-q)
S3n=a1(1-q^3n)/(1-q)
所以(S2n-Sn)²=[a1/(1-q)]²(1-q^2n-1+q^n)²
S3n-S2n=[a1/(1-q)](1-q^3n-1+q^2n)
即证明(1-q^2n-1+q^n)²=(1-q^n)(1-q^3n-1+q^2n)
(1-q^2n-1+q^n)²
=(q^2n-q^n)²
=q^2n(1-q^n)
(1-q^n)(1-q^3n-1+q^2n)
=(1-q^n)(q^2n-q^3n)
=(1-q^n)*q^2n(1-q^n)
=q^2n(1-q^n)²
所以(S2n-Sn)²=Sn(S3n-S2n)
若q=-1
且n是偶数时
若an是q^2n(1-q^n)²=0
即S2n-Sn=S3n-S2n=0
此时不是等比数列