等比数列an的前N项和为Sn,sn=2,s2n,则s3n=?
问题描述：
等比数列an的前N项和为Sn,sn=2,s2n,则s3n=?
答
这是一类题型：Sn=a≠0,S2n=b,S3n=?
设公比为q,则b=S2n=a1+a2+……+an+a(n+1)+a(n+2)+……+a(2n)=
a1+a2+……+an+a1*q^n+a2*q^n+……+a(n)*q^n=
Sn+q^n*Sn=(1+q^n)Sn=(1+q^n)*a,解得q^n=b/a-1
于是S(3n)=a1+a2+……+an+a1*q^n+a2*q^n+……+a(n)*q^n+a1*q^2n+a2*q^2n+……+a(n)*q^2n
=(1+q^n+q^2n)Sn=[1+b/a-1+(b/a-1)^2]a=b+a(b/a-1)^2=a-b+b^2/a