【答案】
分析:(1)先由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/0.png)
,得出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/1.png)
,又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/2.png)
從而求b
1,b
2,b
3,b
4的值;(2)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/3.png)
兩邊同減去1,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/4.png)
,對(duì)上式取倒數(shù),則數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/5.png)
是以-4為首項(xiàng),-1為公差的等差數(shù)列,最后利用等差數(shù)列的通項(xiàng)公式即可求數(shù)列{b
n}的通項(xiàng)公式;
(3)由(2)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/6.png)
,而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/7.png)
,利用拆項(xiàng)法求得Sn,又因4a•S
n>b
n對(duì)n∈N*恒成立,有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/8.png)
最后利用分離參數(shù)a的方法即可求得實(shí)數(shù)a的取值范圍.
解答:解:(1)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/9.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/10.png)
(*),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/11.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/12.png)
(2)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/13.png)
兩邊同減去1,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/14.png)
對(duì)上式取倒數(shù),得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/15.png)
,又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/16.png)
則數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/17.png)
是以-4為首項(xiàng),-1為公差的等差數(shù)列,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/18.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/19.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/20.png)
(3)由(2)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/21.png)
,而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/22.png)
又S
n=a
1•a
2+a
2•a
3++a
n•a
n+1,則有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/23.png)
又因4a•S
n>b
n對(duì)n∈N*恒成立,則有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/24.png)
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/25.png)
對(duì)n∈N*恒成立.
設(shè)函數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/26.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/27.png)
所以g(n)是單調(diào)遞減,則當(dāng)n=1時(shí),g(n)取得最大值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/28.png)
∴4a>4+11即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/29.png)
所以實(shí)數(shù)a的取值范圍為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125004834586216/SYS201310251250048345862010_DA/30.png)
.
點(diǎn)評(píng):本小題主要考查數(shù)列遞推式、數(shù)列與不等式的綜合、不等式的性質(zhì)等基礎(chǔ)知識(shí),考查運(yùn)算求解能力、化歸與轉(zhuǎn)化思想.屬于基礎(chǔ)題.