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
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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.

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