题目内容

已知数列{an}和{bn}满足:a=1,a1=2,a2>0,bn=
a1an+1
(n∈N*)
.且{bn}是以
a为公比的等比数列.
(Ⅰ)证明:aa+2=a1a2
(Ⅱ)若a3n-1+2a2,证明数例{cx}是等比数例;
(Ⅲ)求和:
1
a1
+
1
a2
+
1
a3
+
1
a4
+
+
1
a2n-1
+
1
a2n
分析:(Ⅰ)由
bn+1
bn
=q
,知
an+1an+2
anan+1
=
an+2
an
=q
,由此可得an+2=anq2(n∈N*).
(Ⅱ)由题意知a2n-1=a1q2n-2,a2n=a2qn-2,所以cn=a2n-1+2a2n=5q2n-2.由此可知{cn}是首项为5,以q2为公比的等比数列.
(Ⅲ)由题设条件得
1
a2n-1
=
1
a1
q2-2n
1
a2n
=
1
a2
q2-2n
,所以
1
a1
+
1
a2
+…+
1
a2n
=(
1
a1
+
1
a3
+…+
1
a2n-1
)+(
1
a2
+
1
a4
+…+
1
a2n
)
=
3
2
(1+
1
q2
+
1
q1
+…+
1
q2n-2
)
.由此可知
1
a1
+
1
a2
+…+
1
a2n
=
3
2
n,q=1
3
2
[
q2n-1
q2n-2(q2-1)
],q≠1.
解答:解:(Ⅰ)证:由
bn+1
bn
=q

an+1an+2
anan+1
=
an+2
an
=q

∴an+2=anq2(n∈N*).

(Ⅱ)证:∵an=qn-2q2
∴a2n-1=a2n-3q2=a1q2n-2
a2n=a2n-2q2=a2qn-2
∴cn=a2n-1+2a2n=a1q2n-2+2a2q2n-2=(a1+2a2)q2n-2=5q2n-2
∴{cn}是首项为5,以q2为公比的等比数列.

(Ⅲ)由(Ⅱ)得
1
a2n-1
=
1
a1
q2-2n
1
a2n
=
1
a2
q2-2n
,于是
1
a1
+
1
a2
+…+
1
a2n

=(
1
a1
+
1
a3
+…+
1
a2n-1
)+(
1
a2
+
1
a4
+…+
1
a2n
)

=
1
a1
(1+
1
q2
+
1
q4
+…+
1
q2n-2
)+
1
a2
(1+
1
q2
+
1
q4
+…+
1
q2n-2
)

=
3
2
(1+
1
q2
+
1
q1
+…+
1
q2n-2
)

当q=1时,
1
a1
+
1
a2
+…+
1
a2n
=
3
2
(1+
1
q2
+
1
q4
+…+
1
q2n-2
)
=
3
2
n

当q≠1时,
1
a1
+
1
a2
+…+
1
a2n
=
3
2
(1+
1
q2
+
1
q4
+…+
1
q2n-2
)
=
3
2
(
1-q-2n
1-q-2
)
=
3
2
[
q2n-1
q2n-2(q2-1)
]

1
a1
+
1
a2
+…+
1
a2n
=
3
2
n,q=1
3
2
[
q2n-1
q2n-2(q2-1)
],q≠1.
点评:本题主要考查等比数列的定义,通项公式和求和公式等基本知识及基本的运算技能,考查分析问题能力和推理能力.
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