已知等比数列an的公比q≠-1,前n项和为Sn,求证Sn,S2n-n,S3n-2n成等比数列

Question

已知等比数列an的公比q≠-1,前n项和为Sn,求证Sn,S2n-n,S3n-2n成等比数列

Answer 1

等比数列前n项和为
Sn=a1+a2+a3+.+an
S2n-Sn=a(n+1)+a(n+2)+a(n+3)+.+a2n
=a1*q^n+a2*q^n+a3*q^n+.+an*q^n
=(q^n)*(a1+a2+a3+.+an)
=Sn*q^n
S3n-S2n =a(2n+1)+a(2n+2)+a(2n+3)+.+a3n
=a1 *q^2n+a2*q^2n+a3*q^2n+.+an*q^2n
=(q^2n)（a1+a2+a3+.+an）
=Sn*q^2n
Sn*(S3n-S2n)=(Sn^2)*(q^2n)
(S2n-Sn）²=（ Sn^2)*(q^2n)
Sn*S3n-S2n=（S2n-Sn）²,所以Sn,S2n-Sn,S3n-S2n成等比数列,公比为q^n.