已知数列{an}的通项为an,前n项的和为Sn,且有Sn=2-3an.(1)求an;(2)求数列{nan}的前n项和.-数学

题目简介

已知数列{an}的通项为an,前n项的和为Sn,且有Sn=2-3an.(1)求an;(2)求数列{nan}的前n项和.-数学

题目详情

已知数列{an}的通项为an,前n项的和为Sn,且有Sn=2-3an
(1)求an
(2)求数列{nan}的前n项和.
题型:解答题难度:中档来源:不详

答案

(1)n=1时,s1=2-3a1
∴a1=class="stub"1
2

当n≥2时3an=2-Sn①
3an-1=2-Sn-1②
①-②得   3(an-an-1)=-an,
4an=3an-1
an
an-1
=class="stub"3
4

∵{an}是公比为class="stub"3
4
,首项为class="stub"1
2
的等比数列,an=class="stub"1
2
(class="stub"3
4
)n-1

(2)∵an=class="stub"1
2
(class="stub"3
4
)n-1=class="stub"2
3
•class="stub"3
4
(class="stub"3
4
)n-1=class="stub"2
3
•(class="stub"3
4
)n

Tn=class="stub"2
3
[1•(class="stub"3
4
)+2•(class="stub"3
4
)2+…+n•(class="stub"3
4
)n]①

class="stub"3
4
Tn=class="stub"2
3
[1•(class="stub"3
4
)2+2•(class="stub"3
4
)3+…+n•(class="stub"3
4
)n+1]②

①-②得   class="stub"1
4
Tn=class="stub"2
3
[1•(class="stub"3
4
)+(class="stub"3
4
)2+…+(class="stub"3
4
)n-n•(class="stub"3
4
)n+1]

Tn=class="stub"8
3
[
class="stub"3
4
[1-(class="stub"3
4
)
n
]
1-class="stub"3
4
-n•(class="stub"3
4
)n+1]=8[1-(class="stub"3
4
)n]-class="stub"8
3
n•(class="stub"3
4
)n+1

=8-8(class="stub"3
4
)n-class="stub"8
3
n(class="stub"3
4
)n+1=8-(class="stub"3
4
)n[8+class="stub"8
3
n•class="stub"3
4
]=8-(class="stub"3
4
)n(8+2n)

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