【答案】
分析:(1)当n=1时,S
1=b
1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/0.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/1.png)
=b
1,原式成立.假设当n=k时,S
k=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/2.png)
成立,由此证明n=k+1时,等式仍然成立.
(2)由3a
5=8a
12>0,知3a
5=8(a
5+7d),a
5=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/3.png)
>0,所以d<0.由a
16=a
5+11d=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/4.png)
>0,a
17=a
5+12d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/5.png)
<0,知a
1>a
2>a
3>…>a
16>0>a
17>a
18,b
1>b
2>b
3>…>b
14>0>b
17>b
18,由此能够推导出S
n中S
16最大.
解答:解:(1)证明:当n=1时,S
1=b
1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/7.png)
=b
1,原式成立.(1分)
假设当n=k时,S
k=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/8.png)
成立,(2分)
则S
k+1=S
k+b
k+1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/9.png)
(4分)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/13.png)
(6分)
所以n=k+1时,等式仍然成立,故对于任意n∈N*,都有S
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/14.png)
;(8分)
(2)因为3a
5=8a
12>0,所以3a
5=8(a
5+7d),a
5=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/15.png)
>0,所以d<0
又a
16=a
5+11d=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/16.png)
>0,a
17=a
5+12d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/17.png)
<0,(11分)
所以a
1>a
2>a
3>…>a
16>0>a
17>a
18,b
1>b
2>b
3>…>b
14>0>b
17>b
18,
因为b
15=a
15a
16a
17<0,b
16=a
16a
17a
18>0,(13分)
a
15=a
5+10d=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/18.png)
>0,a
18=a
5+13d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024182547523220543/SYS201310241825475232205021_DA/19.png)
<0,
所以a
15<-a
18,所以b
15>-b
16,b
15+b
16>0,(15分)
故S
16>S
14,所以S
n中S
16最大.(16分)
点评:本题考查数列和函数的综合运用,解题时要认真审题,注意数列归纳法的合理运用,恰当地进行等价转化.