【答案】
分析:根據(jù)題目給出的遞推式,取n=n+1時(shí)得到另外一個(gè)式子,兩式作差后兩邊平方運(yùn)算,得到
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/0.png)
,構(gòu)造數(shù)列設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/1.png)
,則數(shù)列{b
n}為等差數(shù)列,寫出等差數(shù)列的通項(xiàng)公式,把b
n代入后可求a
n,結(jié)合
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/2.png)
可對(duì)求出的a
n進(jìn)行取舍.
解答:解:∵2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/3.png)
①
∴2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/4.png)
②
②-①得:2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/5.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/6.png)
,
兩邊平方得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/7.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/8.png)
設(shè)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/9.png)
,則b
n+1-b
n=4,
而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/10.png)
.
所以數(shù)列{b
n}是首項(xiàng)為2,公差為4的等差數(shù)列,b
n=2+4(n-1)=4n-2.
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/11.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/12.png)
,又a
n>0>0,故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/13.png)
,
從而
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/14.png)
,解得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/15.png)
,
而a
1=1,由2(a
1+a
2)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/16.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/17.png)
,解得a
2=-1±
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/18.png)
,
取
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/19.png)
-1>0,則只有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/20.png)
符合.
所以,此數(shù)列的通項(xiàng)公式
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/21.png)
.
故答案為有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103103013535763344/SYS201311031030135357633013_DA/22.png)
(n∈N
*).
點(diǎn)評(píng):本題考查了數(shù)列的概念及簡(jiǎn)單表示法,考查了利用遞推式求數(shù)列的通項(xiàng)公式,在遞推式中替換n=n+1(或n-1)得另外一個(gè)遞推式,兩式聯(lián)立求解是解答此類問(wèn)題常用的方法,解答該題的關(guān)鍵是兩式作差后兩邊平方,然后構(gòu)造函數(shù),這也是該題的難點(diǎn)所在,該題是中檔題.