设数列{an}满足a1+3a2+3^2×a3+……+3^(n-1)×an=n/3,a∈N (1)求

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设数列{an}满足a1+3a2+3^2×a3+……+3^(n-1)×an=n/3,a∈N (1)求数列{an}的通项(2)设bn=n/an,求数列{bn}的前n项和Sn

$设数列{an}满足a1+3a2+3^2×a3+……+3^(n-1)×an=n/3,a∈N (1)求$

由a1+3a2+3^2a3+……+3^(n-1)an=n/3
和a1+3a2+3^2a3+……+3^(n-1)an+3^na_(n+1)=(n+1)/3得
3^n*a_(n+1)=1/3
所以a_(n+1)=1/[3^(n+1)]
所以an=1/(3^n)=
所以bn=n*3^n
设它的前n项和为S
则S=3+2*3^2+…………n*3^n
3S=3^2+2*3^3+…………(n-1)*3^n+n*3^(n+1)
上两等式左右分别相减得
(1-3)S=3+3^2+3^3+…………3^n-3^(n+1)
=[3^(n+1)-3]/2+3^n-3^(n+1)
=3^n-[3^(n+1)+3]/2
所以S=[3^(n+1)+3]-2*3^n

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