已知等比数列{an}的公比为q,首项为a1,其前n项的和为Sn.数列{an2}的前n项的和为An,数列{(-1)n+1an}的前n项的和为Bn.
(1)若A2=5,B2=-1,求{an}的通项公式;
(2)①当n为奇数时,比较BnSn与An的大小;
②当n为偶数时,若|q|≠1,问是否存在常数λ(与n无关),使得等式(Bn-λ)Sn+An=0恒成立,若存在,求出λ的值;若不存在,说明理由.
【答案】
分析:(1)由题意知
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/0.png)
,由此可知
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/1.png)
,或a
n=2
n-1.
(2)由题设条件知数列{a
n2},{(-1)
n+1a
n}均为等比数列,首项分别为a
12,a
1,公比分别为q
2,-q.
①当n为奇数时,当q=1时,B
nS
n=na
12=A
n.当q=-1时,B
nS
n=na
12=A
n.当q≠±1时,B
2k-1S
2k-1=A
2k-1.综上所述,当n为奇数时,B
nS
n=A
n.
②当n为偶数时,存在常数
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/2.png)
,使得等式(B
n-λ)S
n+A
n=0恒成立.由此入手能够推导出存在常数
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/3.png)
,使得等式(B
n-λ)S
n+A
n=0恒成立.
解答:解:(1)∵A
2=5,B
2=-1,
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/4.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/5.png)
或
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/6.png)
(2分)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/7.png)
,或a
n=2
n-1.(4分)
(2)∵
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/8.png)
=常数,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/9.png)
=常数,
∴数列{a
n2},{(-1)
n+1a
n}均为等比数列,
首项分别为a
12,a
1,公比分别为q
2,-q.(6分)
①当n为奇数时,当q=1时,S
n=na
1,A
n=na
12,B
n=a
1,
∴B
nS
n=na
12=A
n.当q=-1时,S
n=a
1,A
n=na
12,B
n=na
1,
∴B
nS
n=na
12=A
n.(8分)
当q≠±1时,设n=2k-1(k∈N
*),
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/10.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/11.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/12.png)
,
∴B
2k-1S
2k-1=A
2k-1.综上所述,当n为奇数时,B
nS
n=A
n.(10分)
②当n为偶数时,存在常数
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/13.png)
,
使得等式(B
n-λ)S
n+A
n=0恒成立.(11分)
∵|q|≠1,∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/14.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/15.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/16.png)
.
∴(B
n-λ)S
n+A
n=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/17.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/18.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/19.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/20.png)
.(14分)
由题设,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/21.png)
对所有的偶数n恒成立,
又
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/22.png)
,∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/23.png)
.(16分)
∴存在常数
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131025125723213128126/SYS201310251257232131281005_DA/24.png)
,使得等式(B
n-λ)S
n+A
n=0恒成立.
点评:本题考查数列的性质和应用,解题时要认真审题,仔细解答,避免出错.