【答案】
分析:解:(1)由f(x)=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/0.png)
结合b
n=f
-1(n)若对于任意n∈N
*都有b
n=a
n求解,
(2)由正整数c
n的前n项和
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/1.png)
则由通项与前n项和之间的关系求解,要注意分类讨论;
(3)在(1)和(2)的条件下,d
1=2,∴D
1=2,则n≥2时,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/2.png)
,由D
n是数列d
n的前n项和有D
n=
1+d
2+…+d
n用裂项相消法求解
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/3.png)
,再由D
n>log
a(1-2a)恒成立,即log
a(1-2a)小于D
n的最小值,只要求得D
n的最小值即可.
解答:解:(1)由题意得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/4.png)
∵
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/5.png)
∴P=-1∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/6.png)
(2)∵正整数c
n的前n项和
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/7.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/8.png)
解之得∴c
1=1,s
1=1
当n≥2时,c
n=s
n-s
n-1∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/9.png)
∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/10.png)
s
n2-s
n-12=n
∴s
n-12-s
n-22=n-1
s
n-22-s
n-22=n-2
s
22-s
12=2
以上各式累加,得∴
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/11.png)
,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/12.png)
(3)在(1)和(2)的条件下,d
1=2∴D
1=2
当n≥2时,设
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/13.png)
,由D
n是数列d
n的前n项和
有D
n=
1+d
2+…+d
n=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/14.png)
=
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/15.png)
综上
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/16.png)
因为D
n>log
a(1-2a)恒成立,所以log
a(1-2a)小于D
n的最小值,
显然D
n的最小值在n=1时取得,即[D
n]
min=2
∴log
a(1-2a)<2
∴a满足的条件是
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131024182727515385595/SYS201310241827275153855021_DA/17.png)
,∴log
a(1-2a)<2
解得
点评:本题一道新定义题,考查了反函数的求法,数列通项与前n项和间的关系以及累加法求通项和裂项相消法求前n项和等知识和方法,综合性较强.