已知數(shù)列{an}的前n項的和為Sn,且Sn=2n+7-2an.
(1)求證:{an-2}為等比數(shù)列;
(2)是否存在實數(shù)k,使得an≤n3+kn2+9n對于任意的n∈N*都成立,若存在,求出實數(shù)k的取值范圍;若不存在,說明理由.
【答案】
分析:(1)由n=1,解得a
1=3.由n≥2,得3a
n=2a
n-1+2,故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/0.png)
,由此能夠證明{a
n-2}是首項為1,公比為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/1.png)
的等比數(shù)列.
(2)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/2.png)
,知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/3.png)
,由2+(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/4.png)
)
n-1≤n
3+kn
2+9n,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/5.png)
.故只需求出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/6.png)
的最大值即可得到k范圍.
解答:解:(1)n=1時,a
1=S
1=2+7-2a
1,解得a
1=3.
n≥2時,a
n=S
n-S
n-1=2-2a
n+2a
n-1,
即3a
n=2a
n-1+2,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/7.png)
,
∴{a
n-2}是首項為1,公比為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/8.png)
的等比數(shù)列.
(2)由(1)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/9.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/10.png)
,
由2+(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/11.png)
)
n-1≤n
3+kn
2+9n,
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/12.png)
.
∴只需求出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/13.png)
的最大值即可.
設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/15.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/16.png)
,
∵n∈N
*,∴f(n)單調(diào)遞減.
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/18.png)
,
∴g(n)<g(n+1),
故g(n)單調(diào)遞減.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/20.png)
,
當n≥3時,h(n)>h(n+1),
故n≥3時,h(n)單調(diào)遞減.
∴n≥3時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/21.png)
隨著n的增大而減小,
∵p(1)=-7,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/22.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/23.png)
,
∴p(n)的最大值為p(3)=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/24.png)
.
故k≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024184639997791230/SYS201310241846399977912019_DA/25.png)
.
點評:本題考查等比數(shù)列的證明和數(shù)列與不等式的綜合運用,考查運算求解能力,推理論證能力;考查化歸與轉(zhuǎn)化思想.對數(shù)學(xué)思維的要求比較高,有一定的探索性.綜合性強,難度大,是高考的重點,易錯點是判斷最大值時因解題能力差導(dǎo)致失誤.解題時要認真審題,仔細解答,注意提高解題能力.