解:(1)由题意的:f
-1(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834704316.gif)
= f(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834423343.gif)
,所以p =-1,…………2分
所以a
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834579327.gif)
……………………………………………………………………3分翰林汇
(2)因为正数数列{c
n}的前n项之和S
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834454225.gif)
(c
n+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834470350.gif)
),
所以c
1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834454225.gif)
(c
1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834891237.gif)
),解之得:c
1=1,S
1=1……………………………………4分
当n ≥ 2时,c
n = S
n–S
n–1,所以2S
n = S
n–S
n–1 +
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834922446.gif)
,……………………5分
S
n +S
n–1 =
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834922446.gif)
,即:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834954422.gif)
= n,……………………………………7分
所以,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834985435.gif)
= n–1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161835047437.gif)
= n–2,……,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161835078299.gif)
=2,累加得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161835390404.gif)
=2+3+4+……+ n,………………………………………………9分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161835593334.gif)
=1+2+3+4+……+ n =
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161835671485.gif)
,
S
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834610557.gif)
………………………………………………………………10分
(3)在(1)和(2)的条件下,d
1=2,
当n≥2时,设d
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834486426.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161835952468.gif)
=2(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161836077339.gif)
),…………………13分
由D
n是{d
n}的前n项之和,
D
n=d
1+d
2+……+d
n=2[1+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161836170279.gif)
)+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161836217291.gif)
)+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161836233283.gif)
)+……+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161836077339.gif)
)]
=2(2–
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161836280232.gif)
)………………………………………………………………………………16分
因为D
n>log
a (1–2a)恒成立,即log
a (1–2a)恒小于D
n的最小值,
显然D
n的最小值是在n=1时取得,即(D
n)
min=2,
所以log
a (1–2a)<2,1–2a>0,所以0<a<
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823161834688225.gif)
–1…………………………………18分