已知數(shù)列{an}的前n項和Sn=2n2+2n,數(shù)列{bn}的前n項和Tn=2-bn
(Ⅰ)求數(shù)列{an}與{bn}的通項公式;
(Ⅱ)設(shè)cn=an2•bn,證明:當(dāng)且僅當(dāng)n≥3時,cn+1<cn.
【答案】
分析:(1)由題意知a
1=S
1=4,a
n=S
n-S
n-1化簡可得,a
n=4n,n∈N
*,再由b
n=T
n-T
n-1=(2-b
n)-(2-b
n-1),可得2b
n=b
n-1知數(shù)列b
n是等比數(shù)列,其首項為1,公比為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/0.png)
的等比數(shù)列,由此可知數(shù)列{a
n}與{b
n}的通項公式.
(2)由題意知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/1.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/3.png)
.由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/4.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/5.png)
,解得n≥3.由此能夠?qū)С霎?dāng)且僅當(dāng)n≥3時c
n+1<c
n.
解答:解:(1)由于a
1=S
1=4
當(dāng)n≥2時,a
n=S
n-S
n-1=(2n
2+2n)-[2(n-1)
2+2(n-1)]=4n,∴a
n=4n,n∈N
*,
又當(dāng)x≥n時,Tn=2-b
n,∴b
n=2-T
n,
b
n=T
n-T
n-1=(2-b
n)-(2-b
n-1),∴2b
n=b
n-1∴數(shù)列b
n是等比數(shù)列,其首項為1,公比為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/6.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/7.png)
.
(2)由(1)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/8.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/9.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/10.png)
.
由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/11.png)
得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/12.png)
,解得n≥3.
又n≥3時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/13.png)
成立,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/14.png)
,由于c
n>0恒成立.
因此,當(dāng)且僅當(dāng)n≥3時c
n+1<c
n.
點評:由
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023213121691719481/SYS201310232131216917194018_DA/15.png)
可求出b
n和a
n,這是數(shù)列中求通項的常用方法之一,在求出b
n和a
n后,進而得到c
n,接下來用作差法來比較大小,這也是一常用方法.