设数列an满足a1+3a2=3^2 a3+.3^n-1 an=n/3,n属于N,求通项,设bn=n/an,求其Sn

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设数列an满足a1+3a2=3^2 a3+.3^n-1 an=n/3,n属于N,求通项,设bn=n/an,求其Sn
设数列an满足a1+3a2=3^2 a3+.3^n-1 an=n/3,n属于N,求通项,设bn=n/an,求其Sn

设数列an满足a1+3a2=3^2 a3+.3^n-1 an=n/3,n属于N,求通项,设bn=n/an,求其Sn
a1+3a2=3^2 a3+.3^n-1 an=n/3 (1)
a1+3a2=3^2 a3+.3^n-2 a(n-1)=(n-1)/3 (2)
(1)-(2)可得：3^n-1 an=1/3,an=(1/3)^n (错位相减法)
bn=n*3^n
sn=3+18+.n*3^n (3)
s(n-1)=3+18+.(n-1)*3^(n-1) (4)
(3)-3*（4）=-2(3)+3*n*3^n=3+9+.3^n=3*(3^n-1)/2
sn=(3)=(3*n*3^n-3*(3^n-1)/2)/2

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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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