问题描述：
已知数列{an}满足a1,a2-a1,a3-a1,...,an-an-1是首项为1,公比为1/3的等比数列
（1）求{an}的通项公式
（2）如果bn=(2n-1)an,求{bn}的前n项和Sn 主要是第二问 第一问可不答
打错了 题目中应该是a3-a2
问题解答：
我来补答
a(1)=1,
a(n+1)-a(n)=(1/3)^n,
3^na(n+1) = 3*3^(n-1)a(n) + 1,
3^na(n+1) + 1/2 = 3[3^(n-1)a(n) + 1/2]
{3^(n-1)a(n) + 1/2}是首项为a(1)+1/2=3/2,公比为3的等比数列.
3^(n-1)a(n) + 1/2 = (3/2)3^(n-1) = (1/2)3^n,
3^(n-1)a(n) = [3^n - 1]/2,
a(n) = [3 - 1/3^(n-1)]/2.
b(n) = (2n-1)a(n) = (2n-1)[3 - 1/3^(n-1)]/2 = (3/2)(2n-1) - [(2n-1)/2](1/3)^(n-1),
s(n) = b(1)+b(2)+b(3)+...+b(n-1)+b(n)
= (3/2)[2*1-1 + 2*2-1 + 2*3-1 + ...+ 2(n-1)-1 + 2n-1] - [(2*1-1)/2] - [(2*2-1)/2](1/3) - [(2*3-1)/2](1/3)^2 - ...- [(2n-3)/2](1/3)^(n-2) - [(2n-1)/2](1/3)^(n-1),
3s(n) = (9/2)[2*1-1+2*2-1+2*3-1+...+2n-1] - 3[(2*1-1)/2] - [(2*2-1)/2] - [(2*3-1)/2](1/3) - ...- [(2n-3)/2](1/3)^(n-3) - [(2n-1)/2](1/3)^(n-2),
2s(n) = 3s(n)-s(n) = 3[2*1-1+2*2-1+...+2n-1] - 3[(2*1-1)/2] - [2/2] - [2/2](1/3) - ...- [2/2](1/3)^(n-2) + [(2n-1)/2](1/3)^(n-1)
= 3[n(n+1)-n] - 3/2 - [1 + 1/3 + ...+ (1/3)^(n-2)] + [(2n-1)/2](1/3)^(n-1)
= 3n^2 - 3/2 - [1 - (1/3)^(n-1)]/(1-1/3) + [(2n-1)/2](1/3)^(n-1)
= 3n^2 - 3 + (3/2)(1/3)^(n-1) + [(2n-1)/2](1/3)^(n-1)
= 3n^2 - 3 + (n+1)(1/3)^(n-1),
s(n) = (1/2)[3n^2 - 3 + (n+1)(1/3)^(n-1)]
