已知数列{an}满足:a1=12,an=(1-1n+1)an+1+n3n(1)设bn=ann,求数列{bn}的通项公式;(2)求数列{an}的前n项和.-数学

题目简介

已知数列{an}满足:a1=12,an=(1-1n+1)an+1+n3n(1)设bn=ann,求数列{bn}的通项公式;(2)求数列{an}的前n项和.-数学

题目详情

已知数列{an}满足:a1=
1
2
an=(1-
1
n+1
)an+1+
n
3n

(1)设bn=
an
n
,求数列{bn}的通项公式;
(2)求数列{an}的前n项和.
题型:解答题难度:中档来源:不详

答案

(1)∵a1=class="stub"1
2
an=(1-class="stub"1
n+1
)an+1+class="stub"n
3n

由已知有
an+1
n+1
=
an
n
-class="stub"1
3n
bn+1-bn=-class="stub"1
3n

利用累差迭加即可求出数列{bn}的通项公式:bn=class="stub"1
2•3n-1
(n∈N*,n≥2)
经验证知上式对n=1时也成立,
(II)由(I)知an=class="stub"n
2•3n-1
=class="stub"3
2
•class="stub"n
3n
,∴Sn=class="stub"3
2
(class="stub"1
3
+class="stub"2
32
++class="stub"n
3n
)
=class="stub"9
8
-class="stub"9+6n
8•3n

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