题目内容
在等比数列{an}中,an>0(n∈N*),公比q∈(0,1),且a1a5+2a3a5+a2a8=25,又a3与a5的等比中项为2.(1)求数列{an}的通项公式;
(2)设bn=log2an,数列{bn}的前n项和为Sn,求数列{Sn}的通项公式;
(3)是否存在k∈N*,使得
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【答案】分析:(1)根据等比数列的性质可知a1a5=a32,a2a8=a52化简a1a5+2a3a5+a2a8=25得到a3+a5=5,又因为a3与a5的等比中项为2,联立求得a3与a5的值,求出公比和首项即可得到数列的通项公式;
(2)把an代入到bn=
中得到bn的通项公式,即可得到前n项和的通项sn;
(3)把sn代入得到
,讨论求出
各项和的最大值,即可求出k的取值范围.
解答:解:(1)∵a1a5+2a3a5+a2a8=25,
∴a32+2a3a5+a52=25,
∴(a3+a5)2=25,
又an>0,∴a3+a5=5,
又a3与a5的等比中项为2,
∴a3a5=4.
而q∈(0,1),
∴a3>a5,∴a3=4,a5=1,
∴q=
,a1=16,∴an=16×(
)n-1=25-n.
(2)∵bn=log2an=5-n,∴bn+1-bn=-1,
b1=log2a1=log216=log224=4,
∴{bn}是以b1=4为首项,-1为公差的等差数列,
∴Sn=
.
(3)由(2)知Sn=
,∴
=
.
当n≤8时,
>0;当n=9时,
=0;
当n>9时,
<0.
∴当n=8或9时,
+
+
++
=18最大.
故存在k∈N*,使得
+
++
<k对任意n∈N*恒成立,k的最小值为19.
点评:考查学生灵活运用等比数列性质的能力,掌握等比数列的通项公式,会进行数列的求和,理解函数恒成立时所取的条件.
(2)把an代入到bn=
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(3)把sn代入得到
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解答:解:(1)∵a1a5+2a3a5+a2a8=25,
∴a32+2a3a5+a52=25,
∴(a3+a5)2=25,
又an>0,∴a3+a5=5,
又a3与a5的等比中项为2,
∴a3a5=4.
而q∈(0,1),
∴a3>a5,∴a3=4,a5=1,
∴q=
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(2)∵bn=log2an=5-n,∴bn+1-bn=-1,
b1=log2a1=log216=log224=4,
∴{bn}是以b1=4为首项,-1为公差的等差数列,
∴Sn=
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(3)由(2)知Sn=
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当n≤8时,
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当n>9时,
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∴当n=8或9时,
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故存在k∈N*,使得
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点评:考查学生灵活运用等比数列性质的能力,掌握等比数列的通项公式,会进行数列的求和,理解函数恒成立时所取的条件.
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