【答案】
分析:(Ⅰ)把n=1代入2a
n+1+S
n=3,再由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/0.png)
,能求出a
2的值.由2a
n+1+S
n=3,2a
n+S
n-1=3(n≥2)相減,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/1.png)
,由此能夠求出a
n.
(Ⅱ)由題意知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/2.png)
,由此能夠求出滿足條件的所有的n的值.
解答:解:(Ⅰ)由2a
n+1+S
n=3,得2a
2+a
1=3,
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/3.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/4.png)
.
由2a
n+1+S
n=3,2a
n+S
n-1=3(n≥2)相減,
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/5.png)
,
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/6.png)
,所以數(shù)列{a
n}是以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/7.png)
為首項(xiàng),
以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/8.png)
為公比的等比數(shù)列.
因此
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/9.png)
(n∈N
*).
(Ⅱ)由題意與(Ⅰ),
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/10.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/11.png)
因?yàn)?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/12.png">,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125232308706288/SYS201310251252323087062017_DA/13.png)
,
所以n的值為3,4.
點(diǎn)評:本題主要考查數(shù)列遞推關(guān)系,等比數(shù)列的定義,求和公式等基礎(chǔ)知識(shí),同時(shí)考查運(yùn)算求解能力.雖然是一道基礎(chǔ)題,但考查數(shù)列基礎(chǔ)知識(shí)的面比較廣.