【答案】
分析:(I)利用遞推關系可得,n≥2 時,a
n=S
n-S
n-1=4×3
n-1由{a
n}是等比數(shù)列可得a
1=S
1=6+k=4從而苛求得k=-2,代入可求通項公式
(II)結合(I)可求得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/0.png)
,根據(jù)通項公式的特點求和時可利用錯位相減可求T
n,要比較3-16T
n 與
4(n+1)b
n+1 的大小,可通過作差法可得,4(n+1)b
n+1-(3-16T
n)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/1.png)
通過討論n的范圍判斷兩式的大小
解答:解:(Ⅰ)由S
n=2-3
n+k可得
n≥2 時,a
n=S
n-S
n-1=4×3
n-1∵{a
n}是等比數(shù)列
∴a
1=S
1=6+k=4∴k=-2,a
n=4×3
n-1(Ⅱ)由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/2.png)
和a
n=4×3
n-1得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/3.png)
(6分)
T
n=b
1+b
2+…+b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/4.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/5.png)
兩式相減可得,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/6.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/8.png)
4(n+1)b
n+1-(3-16T
n)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/9.png)
而n(n+1)-3(2n+1)=n
2-5n-3
當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/10.png)
或
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/11.png)
<0時,有n(n+1)>3(2n+1)
所以當n>5時有3-16T
n<4(n+1)b
n+1那么同理可得:當
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181210379948831/SYS201310241812103799488020_DA/12.png)
時有n(n+1)<3(2n+1),所以當1≤n≤5時有3-16T
n>4(n+1)b
n+1綜上:當n>5時有3-16T
n<4(n+1)b
n+1;
當1≤n≤5時有3-16T
n>4(n+1)b
n+1點評:本題主要考查了等比數(shù)列的通項公式、由遞推關系求數(shù)列的通項,錯位相減求數(shù)列的和,及通過作差比較大小等知識的綜合應用,屬于綜合試題.