题目内容
已知数列{an}和{bn}满足:a=1,a1=2,a2>0,bn=a1an+1 |
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
=q,知
=
=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
=
q2-2n,
=
q2-2n,所以
+
+…+
=(
+
+…+
)+(
+
+…+
)=
(1+
+
+…+
).由此可知
+
+…+
=
bn+1 |
bn |
| ||
|
| ||
an |
(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 |
1 |
a2n |
1 |
a2 |
1 |
a1 |
1 |
a2 |
1 |
a2n |
1 |
a1 |
1 |
a3 |
1 |
a2n-1 |
1 |
a2 |
1 |
a4 |
1 |
a2n |
3 |
2 |
1 |
q2 |
1 |
q1 |
1 |
q2n-2 |
1 |
a1 |
1 |
a2 |
1 |
a2n |
|
解答:解:(Ⅰ)证:由
=q,
有
=
=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
=
q2-2n,
=
q2-2n,于是
+
+…+
=(
+
+…+
)+(
+
+…+
)
=
(1+
+
+…+
)+
(1+
+
+…+
)
=
(1+
+
+…+
).
当q=1时,
+
+…+
=
(1+
+
+…+
)=
n.
当q≠1时,
+
+…+
=
(1+
+
+…+
)=
(
)=
[
].
故
+
+…+
=
bn+1 |
bn |
有
| ||
|
| ||
an |
∴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 |
1 |
a2n |
1 |
a2 |
1 |
a1 |
1 |
a2 |
1 |
a2n |
=(
1 |
a1 |
1 |
a3 |
1 |
a2n-1 |
1 |
a2 |
1 |
a4 |
1 |
a2n |
=
1 |
a1 |
1 |
q2 |
1 |
q4 |
1 |
q2n-2 |
1 |
a2 |
1 |
q2 |
1 |
q4 |
1 |
q2n-2 |
=
3 |
2 |
1 |
q2 |
1 |
q1 |
1 |
q2n-2 |
当q=1时,
1 |
a1 |
1 |
a2 |
1 |
a2n |
3 |
2 |
1 |
q2 |
1 |
q4 |
1 |
q2n-2 |
3 |
2 |
当q≠1时,
1 |
a1 |
1 |
a2 |
1 |
a2n |
3 |
2 |
1 |
q2 |
1 |
q4 |
1 |
q2n-2 |
3 |
2 |
1-q-2n |
1-q-2 |
3 |
2 |
q2n-1 |
q2n-2(q2-1) |
故
1 |
a1 |
1 |
a2 |
1 |
a2n |
|
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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