【答案】
分析:(1)由已知可求S
2,S
3,进而求出a
2,a
3,结合等差数列的性质可求
(2)由(1)可知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/1.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/2.png)
,利用叠乘法即可求解S
n,然后根据n≥2时,a
n=S
n-S
n-1,n=1时,a
1=S
1,可求
(3)由题意可求h
n,结合数列的通项的特点考虑利用错位相减求解数列的和
解答:解:(1)∵S
1=1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/4.png)
∴S
2=1+c,S
3=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/5.png)
∴a
2=c,a
3=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/6.png)
∵a
1,a
2,a
3成等差数列
∴2a
2=a
1+a
3∴2c=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/7.png)
解可得,c=1(舍)或c=2
(2)∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/10.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/11.png)
则
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/13.png)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/14.png)
∵a
1=S
1=1
当n≥2时,a
n=S
n-S
n-1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/15.png)
=n
当n=1时,a
1=S
1=1也适合上式
故a
n=n
(3)由题意可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/16.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/17.png)
=(-1)
n-1•n•2
n-1=n•(-2)
n-1∴T
n=1•(-2)
+2•(-2)+…+n•(-2)
n-1-2T
n=1•(-2)+2•(-2)
2+…+(n-1)(-2)
n-1+n•(-2)
n两式相减可得,3T
n=1+(-2)+(-2)
2+…+(-2)
n-1-n•(-2)
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131103102924741436690/SYS201311031029247414366021_DA/18.png)
-n•(-2)
n∴
点评:本题主要考查了利用数列的递推公式求解数列的 通项公式,等差数列的性质的应用及叠乘法的应用,还考查了错位相减求解数列的和,是数列知识的综合应用