设数列{an}满足a1+3a2+32a3+…+3n-1an=n3,n∈N*.(1)求数列{an}的通项;(2)设bn=nan,求数列{bn}的前n项和Sn.-数学

题目简介

设数列{an}满足a1+3a2+32a3+…+3n-1an=n3,n∈N*.(1)求数列{an}的通项;(2)设bn=nan,求数列{bn}的前n项和Sn.-数学

题目详情

设数列{an}满足a1+3a2+32a3+…+3n-1an=
n
3
,n∈N*
(1)求数列{an}的通项;
(2)设bn=
n
an
,求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:山东

答案

(1)∵a1+3a2+32a3+…+3n-1an=class="stub"n
3
,①
∴当n≥2时,a1+3a2+32a3+…+3n-2an-1=class="stub"n-1
3
.②
①-②,得3n-1an=class="stub"1
3
an=class="stub"1
3n
(n≥2),
在①中,令n=1,
a1=class="stub"1
3
.∴an=class="stub"1
3n

(2)∵bn=class="stub"n
an

∴bn=n•3n.
∴Sn=3+2×32+3×33+…+n•3n.③
∴3Sn=32+2×33+3×34+…+n•3n+1.④
④-③,得2Sn=n•3n+1-(3+32+33+…+3n),
即2Sn=n•3n+1-
3(1-3n)
1-3

Sn=
(2n-1)3n+1
4
+class="stub"3
4

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