发布网友 发布时间:2024-10-24 00:10
共1个回答
热心网友 时间:2024-11-06 19:30
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^nS3n-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^2nSn*(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.