【答案】
分析:(Ⅰ)推出b
n的表達(dá)式,分別當(dāng)n=1時,求出a
2=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/0.png)
;當(dāng)n=2時,解出a
3=8;
(Ⅱ)設(shè)c
n=a
2n+1-a
2n-1,n∈N
*,利用等比數(shù)列的定義,證明{c
n}是等比數(shù)列;
(Ⅲ)求出S
2n,a
2n,S
2n-1,a
2n-1,求出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/1.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/2.png)
的表達(dá)式,然后求出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/3.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/4.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/5.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/6.png)
的表達(dá)式,利用放縮法證明結(jié)果.
解答:(Ⅰ)解:由b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/7.png)
,(n∈N
*)可得b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/8.png)
又b
n+1a
n+b
na
n+1=(-2)
n+1,
當(dāng)n=1時,a
1+2a
2=-1,可得由a
1=2,a
2=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/9.png)
;
當(dāng)n=2時,2a
2+a
3=5可得a
3=8;
(Ⅱ)證明:對任意n∈N
*,
a
2n-1+2a
2n=-2
2n-1+1…①
2a
2n+a
2n+1=2
2n+1…②
②-①,得a
2n+1-a
2n-1=3×2
2n-1,即:c
n=3×2
2n-1,于是
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/10.png)
所以{c
n}是等比數(shù)列.
(Ⅲ)證明:
a1=2,由(Ⅱ)知,當(dāng)k∈N
*且k≥2時,
a
2k-1=a
1+(a
3-a
1)+(a
5-a
3)+(a
7-a
5)+…+(a
2k-1-a
2k-3)
=2+3(2+2
3+2
5+…+2
2k-3)=2+3×
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/11.png)
=2
2k-1,
故對任意的k∈N
*,a
2k-1=2
2k-1.
由①得2
2k-1+2a
2k=-2
2k-1+1,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/12.png)
k∈N
*,
因此,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/13.png)
于是,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/14.png)
.
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/17.png)
所以,對任意的n∈N
*,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/18.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/19.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/20.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/21.png)
=(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/22.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/23.png)
)+…+(
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/24.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/25.png)
)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/27.png)
=n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/28.png)
≤n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/29.png)
=n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023214336028296420/SYS201310232143360282964019_DA/30.png)
(n∈N
*)
點評:本題考查等比數(shù)列的定義,等比數(shù)列求和等基礎(chǔ)知識,考查計算能力、推理論證能力、綜合發(fā)現(xiàn)問題解決問題的能力以及分類討論思想.