等比数列an的前N项和为Sn,sn=2,s2n,则s3n=?

等比数列an的前N项和为Sn,sn=2,s2n,则s3n=?
等比数列an的前N项和为Sn,sn=2,s2n,则s3n=?

等比数列an的前N项和为Sn,sn=2,s2n,则s3n=?
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

题目不清楚
Sn=a1(1-q^n)/(1-q)
S2n=a1(1-q^2n)/(1-q)
S3n=a1(1-q^3n)/(1-q)
S2n-Sn=a1xq^n(q^n-1)/(q-1)
S3n-S2n=a1xq^(2n)(q^n-1)/(q-1)
可知，Sn,S2n-Sn,S3n-S2n成等比数列
然后你自己算吧