已知数列{an}的前n项和为Sn,a2=32,2Sn+1=3Sn+2(n∈N*).(1)证明数列{an}为等比数列,并求出通项公式;(2)设数列{bn}的通项bn=1an,求数列{bn}的前n项的和T

题目简介

已知数列{an}的前n项和为Sn,a2=32,2Sn+1=3Sn+2(n∈N*).(1)证明数列{an}为等比数列,并求出通项公式;(2)设数列{bn}的通项bn=1an,求数列{bn}的前n项的和T

题目详情

已知数列{an}的前n项和为Sn,a2=
3
2
,2Sn+1=3Sn+2(n∈N*).
(1)证明数列{an}为等比数列,并求出通项公式;
(2)设数列{bn}的通项bn=
1
an
,求数列{bn}的前n项的和Tn
(3)求满足不等式3Tn>Sn(n∈N+)的n的值.
题型:解答题难度:中档来源:不详

答案

(1)由2Sn+1=3Sn+2得到,2Sn=3Sn-1+2(n≥2)
则2an+1=3an(n≥2),
又a2=class="stub"3
2
,2S2=3S1+2,∴a1=1,
a2
a1
=class="stub"3
2

an+1
an
=class="stub"3
2
(n∈N*)

故数列{an}为等比数列,且an=(class="stub"3
2
)n-1

(2)由(1)知,an=(class="stub"3
2
)
n-1
,又由数列{bn}的通项bn=class="stub"1
an
,则bn=(class="stub"2
3
)
n-1

Tn=
1-(class="stub"2
3
)n
1-class="stub"2
3
=3[1-(class="stub"2
3
)n]

(3)由(1)知,an=(class="stub"3
2
)
n-1
,则Sn=
1-(class="stub"3
2
)
n
1-class="stub"3
2
=2[(class="stub"3
2
)n-1]

由(2)知,Tn=3[1-(class="stub"2
3
)
n
]

则3Tn>Sn(n∈N+)⇔9[1-(class="stub"2
3
)
n
]>2[(class="stub"3
2
)
n
-1]

t=(class="stub"3
2
)
n
(t>1),则9(1-class="stub"1
t
)>2(t-1)

解得 1<t<class="stub"9
2
,即1<(class="stub"3
2
)n<class="stub"9
2

又由f(x)=(class="stub"3
2
)x
在R上为增函数,(class="stub"3
2
)3=class="stub"9
2
×class="stub"3
4
(class="stub"3
2
)
4
=class="stub"9
2
×class="stub"9
8

故n=1,2,3

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