【答案】
分析:(Ⅰ)設(shè)等比數(shù)列通式a
n=a
1q
(n-1),根據(jù)S
1>0可知a
1大于零,當(dāng)q不等于1時,根據(jù)s
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/0.png)
>0,進(jìn)而可推知1-q
n>0且1-q>0,或1-q
n<0且1-q<0,進(jìn)而求得q的范圍,當(dāng)q=1時仍滿足條件,進(jìn)而得到答案.
(Ⅱ)把a
n的通項公式代入,可得a
n和b
n的關(guān)系,進(jìn)而可知T
n和S
n的關(guān)系,再根據(jù)(1)中q的范圍來判斷S
n與T
n的大�。�
解答:解:(Ⅰ)設(shè)等比數(shù)列通式a
n=a
1q
(n-1)根據(jù)S
n>0,顯然a
1>0,
當(dāng)q不等于1時,前n項和s
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/1.png)
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/2.png)
>0 所以-1<q<0或0<q<1或q>1
當(dāng)q=1時 仍滿足條件
綜上q>0或-1<q<0
(Ⅱ)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/3.png)
∴b
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/4.png)
=a
nq
2-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/5.png)
a
nq
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/6.png)
a
n(2q
2-3q)
∴T
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/7.png)
(2q
2-3q)S
n
∴T
n-S
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/8.png)
S
n(2q
2-3q-2)=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/9.png)
S
n(q-2)(2q+1)
又因為S
n>0,且-1<q<0或q>0,
所以,當(dāng)-1<q<-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/10.png)
或q>2時,T
n-S
n>0,即T
n>S
n;
當(dāng)-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/11.png)
<q<2且q≠0時,T
n-S
n<0,即T
n<S
n;
當(dāng)q=-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212635834716283/SYS201310232126358347162018_DA/12.png)
,或q=2時,T
n-S
n=0,即T
n=S
n.
點評:本題主要考查了等比數(shù)列的性質(zhì).在解決數(shù)列比較大小的問題上,常利用到不等式的性質(zhì)來解決.