(1)设数列{an}的公比为q,由 a 3 2 =9 a 2 a 6 得 a 23 =9 a 24 , 所以 q 2 = 1 9 , 由条件可知q>0,故 q= 1 3 . 由2a 1 +3a 2 =1得 a 1 = 1 3 . 故数列{a n }的通项式为a n = 1 3 n . (2) b n = n a n =n•3 n , S n =1×3+2× 3 2 +…+n⋅ 3 n , 3 S n =1× 3 2 +2× 3 3 +…+n⋅ 3 n+1 , 两式相减得 -2 S n = 3(1- 3 n ) 1-3 -n⋅ 3 n+1 , 所以 S n = (2n-1)⋅ 3 n+1 +3 4 .