【答案】
分析:(1)根據(jù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/0.png)
,把a
n+b
n=1代入整理得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/1.png)
,進而根據(jù)等差數(shù)列的定義判斷出數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/2.png)
為等差數(shù)列.
(2)根據(jù)a
n+b
n=1,a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/3.png)
求得a
1和b
1.進而根據(jù)(1)中
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/4.png)
求得a
n,進而求得b
n,進而可知要證不等式(1+a
n)
n+1•b
nn>1,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/5.png)
,令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/6.png)
,對函數(shù)f(x)進行求導(dǎo),再令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/7.png)
,對函數(shù)g(x)進行求導(dǎo),進而利用導(dǎo)函數(shù)判斷f(x)和g(x)的單調(diào)性,進而利用函數(shù)的單調(diào)性證明原式.
解答:(1)解:數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/8.png)
為等差數(shù)列.
理由如下:
∵對任意n∈N
*都有a
n+b
n=1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/9.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/10.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/11.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/12.png)
.
∴數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/13.png)
是首項為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/14.png)
,公差為1的等差數(shù)列.
(2)證明:∵a
n+b
n=1,a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/15.png)
∴a
1=b
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/16.png)
.
由(1)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/17.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/19.png)
.
所證不等式(1+a
n)
n+1•b
nn>1,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/20.png)
,
也即證明
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/21.png)
.
令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/22.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/23.png)
.
再令
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/24.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/25.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/26.png)
.
當(dāng)x>1時,g′(x)<0,
∴函數(shù)g(x)在[1,+∞)上單調(diào)遞減.
∴當(dāng)x>1時,g(x)<g(1)=0,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/27.png)
.
∴當(dāng)x>1時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/28.png)
<0.
∴函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/29.png)
在(1,+∞)上單調(diào)遞減.
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/30.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/31.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/32.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/33.png)
.
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834724147/SYS201310232126358347241020_DA/34.png)
.
∴(1+a
n)
n+1•b
nn>1成立.
點評:本小題主要考查導(dǎo)數(shù)及其應(yīng)用、數(shù)列、不等式等知識,考查化歸與轉(zhuǎn)化的數(shù)學(xué)思想方法,以及抽象概括能力、推理論證能力、運算求解能力和創(chuàng)新意識