【答案】
分析:(Ⅰ)由S
n=2a
n+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/0.png)
×(-1)
n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/1.png)
,n=1,2,3,…,再寫(xiě)一式,兩式相減整理可得a
n=2a
n-1+3×(-1)
n-1
(Ⅱ)由(Ⅰ)令b
n=(-1)
na
n得b
n=-2b
n-1-3,構(gòu)造新數(shù)列b
n+1是等比數(shù)列,從而可求數(shù)列{a
n}的通項(xiàng)公式;
(Ⅲ)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/2.png)
,∴S
2k-1=2
2k-1,S
2k=2
2k-1,再進(jìn)行分組求和,利用等比數(shù)列的求和公式可證.
解答:解:(Ⅰ)由S
n=2a
n+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/3.png)
×(-1)
n-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/4.png)
,n=1,2,3,…,①
得S
n-1=2a
n-1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/5.png)
×(-1)
n-1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/6.png)
,n=2,3,…,②…(1分)
將①和②相減得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/7.png)
,n=2,3,…,…(2分)
整理得:a
n=2a
n-1+3×(-1)
n-1,n=2,3,…. …(3分)
(Ⅱ)在已知條件中取n=1得,a
1=2a
1-![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/8.png)
-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/9.png)
,∴a
1═2.…(4分)
∵a
n=2a
n-1+3×(-1)
n-1,∴(-1)
na
n=-2(-1)
n-1a
n-1-3,
∴令b
n=(-1)
na
n得b
n=-2b
n-1-3,n=2,3,….…(5分)
∴b
n+1+1=-2(b
n+1),n=1,2,3,…,
∵b
1+1=-1≠0,∴b
n+1=(-1)×(-2)
n-1,n=1,2,3,…,
∴a
n=2
n-1+(-1)
n-1. …(7分)
(Ⅲ)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/10.png)
,∴S
2k-1=2
2k-1,S
2k=2
2k-1. …(8分)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/11.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/12.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/14.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/15.png)
. …(10分)
同理
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/16.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/17.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/19.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/20.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/21.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/22.png)
<
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182435805771817/SYS201310241824358057718020_DA/23.png)
,n∈N
*. …(12分)
點(diǎn)評(píng):本題主要考查數(shù)列通項(xiàng)公式的求解,考查數(shù)列與不等式的綜合,有一定的難度.