【答案】
分析:(法一)(I)由a
1結(jié)合遞推公式可求a
2,a
3,a
4,代入
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/0.png)
求b
1,b
2,b
3,b
4(II)先由(I)中求出的b
1,b
2,b
3,b
4的值,觀察規(guī)律可猜想數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/1.png)
為等比數(shù)列,進(jìn)而可求b
n,結(jié)合
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/2.png)
⇒
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/3.png)
,從而猜想得以證明,代入求出a
n•b
n,進(jìn)而求出前n和s
n(法二)(I)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/4.png)
代入遞推公式可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/5.png)
,代入可求b
1,b
2,b
3,b
4(II)利用(I)中的遞推關(guān)系個(gè)構(gòu)造數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/6.png)
為等比數(shù)列,從而可求b
n,s
n(法三)(I)同法一
(II)先由(I)中求出的b
1,b
2,b
3,b
4的值,觀察規(guī)律可猜想數(shù)列b
n+1-b
n為等比數(shù)列,仿照法一再證明猜想,根據(jù)求通項(xiàng)的方法求b
n,進(jìn)一步求s
n解答:解:法一:
(I)a
1=1,故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/7.png)
;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/8.png)
,
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/9.png)
;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/10.png)
,
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/11.png)
;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/12.png)
,
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/13.png)
.
(II)因
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/14.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/15.png)
故猜想
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/16.png)
是首項(xiàng)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/17.png)
,公比q=2的等比數(shù)列.
因a
n≠2,(否則將a
n=2代入遞推公式會(huì)導(dǎo)致矛盾)故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/18.png)
.
因
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/19.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/20.png)
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/21.png)
確是公比為q=2的等比數(shù)列.
因
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/22.png)
,故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/23.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/24.png)
,
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/25.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/26.png)
,
故S
n=a
1b
1+a
2b
2+…+a
nb
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/27.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/29.png)
法二:
(Ⅰ)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/30.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/31.png)
,代入遞推關(guān)系8a
n+1a
n-16a
n+1+2a
n+5=0,
整理得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/32.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/33.png)
,
由a
1=1,有b
1=2,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/34.png)
.
(Ⅱ)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/35.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/36.png)
是首項(xiàng)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/37.png)
,公比q=2的等比數(shù)列,
故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/38.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/39.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/40.png)
,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/41.png)
,
故S
n=a
1b
1+a
2b
2+…+a
nb
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/42.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/43.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/44.png)
.
法三:
(Ⅰ)同解法一
(Ⅱ)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/45.png)
猜想{b
n+1-b
n}是首項(xiàng)為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/46.png)
,
公比q=2的等比數(shù)列,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/47.png)
又因a
n≠2,故
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/48.png)
.
因此
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/49.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/50.png)
;
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/51.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/52.png)
.
因
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/53.png)
是公比q=2的等比數(shù)列,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/54.png)
,
從而b
n=(b
n-b
n-1)+(b
n-1-b
n-2)+…+(b
2-b
1)+b
1
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/55.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/56.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/57.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/58.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/59.png)
,
故S
n=a
1b
1+a
2b
2+…+a
nb
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/60.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/61.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181902481178599/SYS201310241819024811785019_DA/62.png)
.
點(diǎn)評(píng):本題考查了數(shù)列的綜合運(yùn)用:遞推關(guān)系的運(yùn)用,構(gòu)造等比求數(shù)列通項(xiàng),累加求通項(xiàng),歸納推理的運(yùn)用,綜合考查了考生的推理運(yùn)算能力.