【答案】
分析:(Ⅰ)2
n+1a
n+1-2
na
n=n,令b
n=2
n+1a
n+1-2
na
n,得2
na
n=2a
1+b
1+b
2+…+b
n-1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/0.png)
,由此能求出數(shù)列{a
n}的通項公式.
(Ⅱ)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/1.png)
,可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/2.png)
,2
n+1=(1+1)
n+1=1+C
n+11+C
n+12+…+C
n+1n-1+C
n+1n+1,所以2
n+1>n
2+2n+2,由此能證明
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/3.png)
.
(Ⅲ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/4.png)
,欲證:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/5.png)
.,即證
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/6.png)
,即ln(1+T
n)-T
n<0.構(gòu)造函數(shù)f(x)=ln(1+x)-x,借助導數(shù)能夠證明
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/7.png)
.
解答:解:(Ⅰ)∵2
n+1a
n+1-2
na
n=n
令b
n=2
n+1a
n+1-2
na
n,∴2
na
n=2a
1+b
1+b
2+…+b
n-1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/8.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/9.png)
,又a
1=1成立∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/10.png)
(4分)
(Ⅱ)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/11.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/12.png)
又當n≥2時,2
n+1=(1+1)
n+1=1+C
n+11+C
n+12+…+C
n+1n-1+C
n+1n+1∴2
n+1>1+C
n+11+2C
n+12,∴2
n+1>n
2+2n+2,而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/13.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/14.png)
,又a
1=1
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/15.png)
(9分)
(Ⅲ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/16.png)
欲證:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/17.png)
.,即證
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/18.png)
,即ln(1+T
n)-T
n<0.
構(gòu)造函數(shù)f(x)=ln(1+x)-x(x≥0),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/19.png)
,
∴f(x)在[0,+∞)上為減函數(shù),f(x)的最大值為f(0)=0,
∴當x>0時,f(x)<0,∴l(xiāng)n(1+T
n)-T
n<0
故不等式
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182049554143074/SYS201310241820495541430021_DA/20.png)
.成立.(14分)
點評:本題考查數(shù)列的性質(zhì)和應(yīng)用,解題時要認真審題,仔細解答,注意構(gòu)造法的合理運用.