在数列an中，设S1=a1+a2+a3,S2=a4+a5+a6,S3=a7+a8+a9.(1）若数列是等比数列，求证：S1，S2，S3成等比数列

解：已知数列an是等比数列
则an=a1q^(n-1)
所以S1=a1+a2+a3=a1(1+q+q^2)
S2=a4+a5+a6=a1(q^3+q^4+q^5)
S3=a7+a8+a9=a1(q^6+q^7+q^8)
因为S2/S1=S3/S2=1/(q^3)
得证S1，S2，S3成等比数列

an=a1*q^(n-1) ---------a4=a1*q^3,a7=a1*q^6
S1=a1(1+q+q^2)
S2=a4(1+q+q^2)=a1*q^3*(1+q+q^2)
S3=a7(1+q+q^2)=a1*q^6*(1+q+q^2)
故S1：S2：S3=1：q^3：q^6
比值Q=q^3

(1)
an=a1q^(n-1)
S1=a1+a2+a3 = a1(1+q+q^2)
S2=a4+a5+a6= a1q^3(1+q+q^2)
S3=a7+a8+a9= a1q^6(1+q+q^2)

S1.S3 =(S2)^2
=> S1,S2,S3 成等比数列
(2)
Assume Sn = a1q^(3n-3)) .(1+q+q^2)

Sn= a(3n-2)+ a(3n-1)+ a(3n)
=a1q^(3n-3) + a1q^(3n-2) + a1q^(3n-1)
= a1q^(3n-3) . ( 1+q+q^2)