【答案】
分析:(1)由題設(shè)知S
n+1=(a
1+1)•4
n-1.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/0.png)
.先證明充分性:當(dāng)a
1=3時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/1.png)
,所以對(duì)n∈N
*,都有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/2.png)
,即數(shù)列{a
n}是等比數(shù)列.再證明必要性:因?yàn)閧a
n}是等比數(shù)列,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/3.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/4.png)
,解得a
1=3.
(2)當(dāng)n=1時(shí),b
1=5+a
1;當(dāng)n≥2時(shí),b
n=5
n-(-1)
n×3(a
1+1)×4
n-2(a
1>-1).當(dāng)n為偶數(shù)時(shí),15(a
1+1)×4
n-2>-4×5
n恒成立.故a
1∈(-1,+∞).當(dāng)n為奇數(shù)時(shí),b
1<b
2且b
n<b
n+1(n≥3)恒成立.5+a
1<25-3(a
1+1),得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/5.png)
.由此入手能夠得到a
1的取值范圍.
解答:解:(1)因?yàn)閿?shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/6.png)
是公比為2的等比數(shù)列,
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/7.png)
,
即S
n+1=(a
1+1)•4
n-1.
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/8.png">所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/9.png)
顯然,當(dāng)n≥2時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/10.png)
.
①充分性:當(dāng)a
1=3時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/11.png)
,所以對(duì)n∈N
*,都有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/12.png)
,即數(shù)列{a
n}是等比數(shù)列.
②必要性:因?yàn)閧a
n}是等比數(shù)列,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/13.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/14.png)
,解得a
1=3.
(2)當(dāng)n=1時(shí),b
1=5+a
1;當(dāng)n≥2時(shí),b
n=5
n-(-1)
n×3(a
1+1)×4
n-2(a
1>-1).
①當(dāng)n為偶數(shù)時(shí),5
n-3(a
1+1)×4
n-2<5
n+1+3(a
1+1)×4
n-1恒成立.
即15(a
1+1)×4
n-2>-4×5
n恒成立.故a
1∈(-1,+∞).
②當(dāng)n為奇數(shù)時(shí),b
1<b
2且b
n<b
n+1(n≥3)恒成立.
由b
1<b
2知,5+a
1<25-3(a
1+1),得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/15.png)
.
由b
n<b
n+1對(duì)n≥3的奇數(shù)恒成立,知5
n+3(a
1+1)×4
n-2<5
n+1-3(a
1+1)×4
n-1恒成立,
即15(a
1+1)×4
n-2<4×5
n恒成立,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/16.png)
恒成立.
因?yàn)楫?dāng)對(duì)n≥3的奇數(shù)時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/17.png)
的最小值為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/18.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/19.png)
.
又因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/20.png">,故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/21.png)
.
綜上所述,b
n<b
n+1對(duì)n∈N
*恒成立時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213201093258942/SYS201310232132010932589004_DA/22.png)
.
點(diǎn)評(píng):本題考查等比數(shù)列的性質(zhì),解題時(shí)感受知識(shí)點(diǎn)的有效組合,注意積累解題方法.