感谢高中数学数列题的分析过程和答案。

所以:an = 1/(3 n)

sn=a1(1-q^n)/(1-q)=(1/3)(1-1/3^n)/(2/3)=(1-1/3^n)/2 =(1-an)/2

(2)bn = log(3)a 1+log(3)a2+…+log(3)an

=log(3)[a1×a2×…×an]

=log(3)[1/3×1/(3^2)×…×1/(3^n)]

=log(3)[1/3^(1+2+…+n)]

=-log(3)[3^(1+2+…+n) ]

=-(1+2+…+n)

=-n(1+n) /2