【答案】
分析:(Ⅰ)由數(shù)列{a
n}為等差數(shù)列,且a
5=14,a
7=20,能得到公差d=3,首項(xiàng)a
1=2.由此能求出{a
n}的通項(xiàng)公式;由數(shù)列{b
n}的前n項(xiàng)和為S
n=1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/0.png)
(n∈N
*),由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/1.png)
,能求出{b
n}的通項(xiàng)公式.
(Ⅱ)由a
n=3n-1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/2.png)
,得c
n=a
n•b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/3.png)
,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/5.png)
,再由錯(cuò)位相減法能求出數(shù)列{c
n}的前n項(xiàng)和T
n.
解答:(Ⅰ)解:∵數(shù)列{a
n}為等差數(shù)列,且a
5=14,a
7=20,
∴公差d=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/6.png)
=3,
∵a
5=a
1+4×3=14,
∴a
1=2.
∴a
n=2+(n-1)×3=3n-1.
∵數(shù)列{b
n}的前n項(xiàng)和為S
n=1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/7.png)
(n∈N
*),
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/8.png)
,
b
n=S
n-S
n-1=[1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/9.png)
]-[1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/10.png)
]=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/11.png)
,
當(dāng)n=1時(shí),
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/13.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/14.png)
.
(Ⅱ)由a
n=3n-1,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/15.png)
,
得c
n=a
n•b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/16.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/17.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/19.png)
T
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/20.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/21.png)
,
兩式相減,得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/23.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124551302723589/SYS201310251245513027235019_DA/24.png)
.
點(diǎn)評(píng):本題考查數(shù)列的通項(xiàng)公式和數(shù)列的前n項(xiàng)和的計(jì)算,綜合性強(qiáng),強(qiáng)難大,是高考的重點(diǎn).解題時(shí)要認(rèn)真審題,注意迭代法和錯(cuò)位相減法的合理運(yùn)用.