已知数列{an}与{bn}有如下关系:a1=2,an+1=12(an+1an),bn=an+1an-1.(1)求数列{bn}的通项公式.(2)设Sn是数列{an}的前n项和,当n≥2时,求证:Sn<n

题目简介

已知数列{an}与{bn}有如下关系:a1=2,an+1=12(an+1an),bn=an+1an-1.(1)求数列{bn}的通项公式.(2)设Sn是数列{an}的前n项和,当n≥2时,求证:Sn<n

题目详情

已知数列{an}与{bn}有如下关系:a1=2,an+1=
1
2
(an+
1
an
),bn=
an+1
an-1

(1)求数列{bn}的通项公式.
(2)设Sn是数列{an}的前n项和,当n≥2时,求证:Sn<n+
4
3
题型:解答题难度:中档来源:不详

答案

(1)∵bn=
an+1
an-1

∴b1=
a1+1
a1-1
=3

∵an+1=class="stub"1
2
(an+class="stub"1
an
),
∴bn+1=
an+1+1
an+1-1
=(
an+1
an-1
)
2
=
b2n
>0

bn=
b2n-1
=…=32n-1

(2)证明:当n≥2时,an+1-1=
an-1
32n-1+1
≤class="stub"1
10
(an-1)

(当且仅当n=2时取等号)且a2=class="stub"1
2
(a1+class="stub"1
a1
)=class="stub"5
4

a3-1≤class="stub"1
10
(a2-1)
a4-1≤class="stub"1
10
(a3-1)
,…,an-1≤class="stub"1
10
(an-1-1)

以上式子累和得Sn-a1-a2-(n-2)≤class="stub"1
10
[Sn-1-a1-(n-2)]

∴10[Sn-a1-a2-(n-2)]≤Sn-1-a1-(n-2)
9Sn≤class="stub"25
2
+9n-
32n-1+1
32n-1-1

Sn≤class="stub"25
18
+n-
32n-1+1
9(32n-1-1)
<class="stub"25
18
+n-class="stub"1
9
=class="stub"23
18
+n<class="stub"24
18
+n

∴Sn<n+class="stub"4
3
.得证

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