题目内容
已知有穷数列{an}共有2k项(整数k≥2),首项a1=2,设该数列的前n项和为Sn,且an+1=(a-1)Sn+2(n=1,2,…,2k-1),其中常数a>1.(1)求证:数列{an}是等比数列;
(2)若a=
,数列{bn}满足bn=
log2(a1a2…an)(n=1,2,…,2k),求数列{bn}的通项公式;
(3)若(2)中的数列{bn}满足不等式.
|b1-
|+|b2-
|+…+|b2k-1-
|+|b2k-
|≤4,求k的值.
解:(1)an+1=(a-1)Sn+2, ①
当n≥2时,an=(a-1)Sn-1+2, ②
两式相减得
an+1-an=(a-1)(Sn-Sn-1)=(a-1)an,∴an+1=aan.
∴
=a为常数.
∴数列{an}是以a1=2为首项,以a为公比的等比数列.
(2)由(1)知an=2·an-1,
∴bn=
log2(2·2a·2a2·…·2an-1)
=
log2(2n·a1+2+…+(n-1))
=
(n+
)=1+
·
·log2a
=1+
·
=1+
.
(3)|bn-
|=|
-
|=|
|,
∴|b1-
|+|b2-
|+…+|b2k-1-
|+|b2k-
|
=|
|+|
|+…+|
|+|
|
=2[
+
+…+
+
]
=
=
.
令
≤4,即k2-8k+4≤0,
∴4-2
≤k≤4+2
.
又∵k≥2,k∈Z,
∴k的值为2,3,4,5,6,7.
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