数列{an}满足a1+3a2+3方a3+···+3的n-1次方an=n/3

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/16 06:53:23

数列{an}满足a1+3a2+3方a3+···+3的n-1次方an=n/3
1.求数列{an}的通项公式
2.设bn=n/an,求数列{bn}的前n项和Sn

a1+3a2+3^2*a3+...+3^(n-1)*an=n/3.(1)
n=1时,a1=1/3,
n>1时,a1+3a2+3^2*a3+...+3^(n-2)*an=(n-1)/3...(2)
(1)-(2),得3^(n-1)*an=1/3,
an=1/(3^n)
n=1时也符合此式,所以通项公式是an=1/(3^n).
bn=n/an=n*3^n.
Sn=1*3^1+2*3^2+...+n*3^n.(3)
3Sn=1*3^2+2*3^3+...+n*3^(n+1).(4)
(4)-(3),得2Sn=-(3^1+3^2+...+3^n)+n*3^(n+1)
=n*3^(n+1)-[3^(n+1)-3]/2
=[(2n-1)*3^(n+1)+3]/2,
Sn=[(2n-1)*3^(n+1)+3]/4.

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