【答案】
分析:(Ⅰ)根據(jù)曲線C:x
2-y
2=1上的點P到點A
n(0,a
n)的距離的最小值為d
n,設(shè)點P(x,y),利用兩點間的距離公式,再采用配方法可得,再根據(jù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/0.png)
,可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/1.png)
,從而可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/2.png)
,從而數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/3.png)
是首項
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/4.png)
,公差為2的等差數(shù)列,進(jìn)而可求數(shù)列{a
n}的通項公式;
(Ⅱ)先判斷a
2n+2a
2n-1<a
2n+1a
2n,從而有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/5.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/6.png)
,疊加可得結(jié)論;
(Ⅲ)先證明
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/7.png)
,從而可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/8.png)
,進(jìn)而可知存在常數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/9.png)
,對?n∈N
*,都有不等式:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/10.png)
成立.
解答:(Ⅰ)解:設(shè)點P(x,y),則x
2-y
2=1,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/11.png)
,
因為y∈R,所以當(dāng)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/12.png)
時,|PA
n|取得最小值d
n,且
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/13.png)
,
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/14.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/15.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/16.png)
將
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/17.png)
代入
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/18.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/19.png)
兩邊平方得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/20.png)
,又a
=0,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/21.png)
故數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/22.png)
是首項
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/23.png)
,公差為2的等差數(shù)列,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/24.png)
,
因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/25.png)
>0,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/26.png)
.…(6分)
(Ⅱ)證明:因為(2n+2)(2n-1)-2n(2n+1)=-2<0,
所以(2n+2)(2n-1)<2n(2n+1)
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/27.png)
,所以a
2n+2a
2n-1<a
2n+1a
2n所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/28.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/29.png)
以上n個不等式相加得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/30.png)
.…(10分)
(Ⅲ)解:因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/31.png)
,當(dāng)k≥2時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/32.png)
,
因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/33.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/34.png)
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/35.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/36.png)
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/37.png)
.
故存在常數(shù)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/38.png)
,對?n∈N
*,都有不等式:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102344468157824/SYS201311031023444681578019_DA/39.png)
成立.…(14分)
點評:本題考查數(shù)列的通項,考查數(shù)列與不等式的綜合,考查放縮法的運用,解題的關(guān)鍵是根據(jù)目標(biāo),適當(dāng)放縮,難度較大.