【答案】
分析:(Ⅰ)要證數(shù)列{
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/0.png)
}是等比數(shù)列;需證
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/1.png)
(n=1,2,3,…)成立,另外應(yīng)說明
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/2.png)
;
(Ⅱ)由(Ⅰ)知數(shù)列{
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/3.png)
}是首項為1,公比為2的等比數(shù)列,可得S
n的通項公式,代入a
n+1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/4.png)
S
n(n=1,2,3,…)可得S
n+1=4a
n.說明當n=1時,S
2=a
1+a
2=4a
1,等式成立.
解答:(I)證:由a
1=1,a
n+1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/5.png)
S
n(n=1,2,3,),
知a
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/6.png)
S
1=3a
1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/7.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/8.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/9.png)
又a
n+1=S
n+1-S
n(n=1,2,3,…),則S
n+1-S
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/10.png)
S
n(n=1,2,3,),
∴nS
n+1=2(n+1)S
n,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/11.png)
(n=1,2,3,…),
故數(shù)列{
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/12.png)
}是首項為1,公比為2的等比數(shù)列.
(II)證明:S
n+1=4a
n.當n=1時,S
2=a
1+a
2=4a
1,等式成立.
由(1)知:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/13.png)
,∴S
n=n2
n-1當n≥2時,4a
n=4(S
n-S
n-1)=2
n(2n-n+1)=(n+1)2
n=S
n+1,等式成立.
因此對于任意正整數(shù)n≥1都有S
n+1=4a
n.
點評:要證一個數(shù)列是等比數(shù)列,利用定義,每一項與它的前一項之比為一個常數(shù),在這兒注意,n=1時,不在其中,所以要加以說明;同樣第二個問題中,a
n+1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024180519277981005/SYS201310241805192779810018_DA/14.png)
S
n(n=1,2,3,…),這個式子也不包括a
1應(yīng)加以說明.