设数列{an},{bn}满足:a1=4,a2=52,an+1=an+bn2,bn+1=2anbnan+bn.(1)用an表示an+1;并证明:∀n∈N+,an>2;(2)证明:{lnan+2an-

题目简介

设数列{an},{bn}满足:a1=4,a2=52,an+1=an+bn2,bn+1=2anbnan+bn.(1)用an表示an+1;并证明:∀n∈N+,an>2;(2)证明:{lnan+2an-

题目详情

设数列{an},{bn}满足:a1=4,a2=
5
2
an+1=
an+bn
2
bn+1=
2anbn
an+bn
.
(1)用an表示an+1;并证明:∀n∈N+,an>2;
(2)证明:{ln
an+2
an-2
}
是等比数列;
(3)设Sn是数列{an}的前n项和,当n≥2时,Sn2(n+
4
3
)
是否有确定的大小关系?若有,加以证明;若没有,请说明理由.
题型:解答题难度:中档来源:济南二模

答案

(1)由已知得a1=4,a2=class="stub"5
2
,所以b1=1故an+1bn+1=anbn═a1b1=4;
由已知:an>0,a1>2,a2>2,bn=class="stub"4
an
an+1=
an
2
+class="stub"2
an

由均值不等式得an+1>2
故∀n∈N+,an>2

(2)
an+1+2
an+1-2
=(
an+2
an-2
)2
an+1+2=
(an+2)2
2an

an+1-2=
(an-2)2
2an

所以ln
an+1+2
an+1-2
=2ln
an+2
an-2
,所以{ln
an+2
an-2
}
是等比数列

(3)由(2)可知ln
an+2
an-2
=(ln3)×2n-1=ln32n-1
an=
32n-1+1
32n-1-1

Cn=class="stub"4
32n-1
=class="stub"4
(32n-2)(32n-2)
<class="stub"1
4
Cn-1
,(n≥2)
Cn<class="stub"1
4
Cn-1<(class="stub"1
4
)2Cn-2<<(class="stub"1
4
)n-1C1=2(class="stub"1
4
)n-1

∴当n≥2时,an<2+2(class="stub"1
4
)n-1

Sn=a1+a2++an<4+2(n-1)+2[class="stub"1
4
+(class="stub"1
4
)
2
++(class="stub"1
4
)
n-1
]

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

=2n+2+class="stub"2
3
(1-class="stub"1
4n-1
)<2n+class="stub"8
3

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