已知Sn是等比数列｛an｝的前n项和,S3,S9,S6成等差数列

a) 当q=1是,S3=3a1,S9=9a1,S6=6a1,则18a1=9a1 a1=0（舍）
q不等于1时,S3=a1(1-q^3)/(1-q),S9=a1(1-q^9)/(1-q),
S6=a1(1-q^6)/(1-q),2S9=S3+S6 有2q^9=q^3+q^6
2q^6=1+q^3 解得q=1(舍),q=-(1/2)^(1/3)
b) an=a1q^(n-1) a(3n-2)=(-1/2)^(n-1)
Tn=1+2(-1/2)+3(-1/2)^2+...+n(-1/2)^(n-1) ①
(-1/2)Tn=0+(-1/2)+2(-1/2)^2+...+(n-1）(-1/2)^(n-1)+n(-1/2)^n ②
①- ②得 3/2Tn=1+(-1/2)+(-1/2)^2+...+(-1/2)^(n-1)-n(-1/2)^n
=(1-(-1/2)(-1/2)^(n-1))/(1-(-1/2))-n(-1/2)^n
=2/3-2/3(-1/2)^n-n(-1/2)^n
所以 Tn=4/9-（4/9+2n/3)(-1/2)^n