已知數(shù)列{an}前n項的和為Sn,前n項的積為Tn,且滿足Tn=2n(1-n).
①求a1;
②求證:數(shù)列{an}是等比數(shù)列;
③是否存在常數(shù)a,使得(Sn+1-a)2=(Sn+2-a)(Sn-a)對n∈N+都成立?若存在,求出a,若不存在,說明理由.
【答案】
分析:(1)由“數(shù)列{a
n}前n項的和為S
n,前n項的積為T
n,且滿足T
n=2
n(1-n)”令n=1可求解.
(2)證明:由T
n=2
n(1-n)解得T
(n-1)=2
(n-1)(2-n)兩式相除,整理可得數(shù)列{a
n}是等比數(shù)列;
(3)由(2)求解得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/0.png)
再求得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/2.png)
代入(S
n+1-a)
2=(S
n+2-a)(S
n-a)兩端驗(yàn)證可即可.
解答:解:(1)∵數(shù)列{a
n}前n項的和為S
n,前n項的積為T
n,且滿足T
n=2
n(1-n).
∴a
1=T
1=2
1(1-1)=1
(2)證明:∵T
n=2
n(1-n).
∴T
(n-1)=2
(n-1)(2-n).
將上面兩式相除,
得:a
n=2
[-2(n-1)].
∴a
n=
(n-1).
∵a
n+1=
(n).
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/5.png)
∴數(shù)列{a
n}是等比數(shù)列;
(3)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/6.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/7.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/8.png)
∵(S
n+1-a)
2=(S
n+2-a)(S
n-a)
∴(S
n+1-a)
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/9.png)
而:(S
n+2-a)(S
n-a)=(S
n+2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/10.png)
)(S
n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/11.png)
)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/12.png)
(S
n+1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/13.png)
)
2=(S
n+2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/14.png)
)(S
n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/15.png)
)對n∈N
+都成立
即:存在常數(shù)a=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213718258593600/SYS201310232137182585936004_DA/16.png)
,使(S
n+1-a)
2=(S
n+2-a)(S
n-a)對n∈N
+都成立.
點(diǎn)評:本題主要考查數(shù)列的類型和數(shù)列的通項公式和前n項和公式,還考查了存在性問題,這類問題一般通過具體的探究出來,再證明.