解:(Ⅰ)∵
![](http://thumb.1010pic.com/pic5/latex/180753.png)
,其中a
1=1,a
n≠0.
∴
![](http://thumb.1010pic.com/pic5/latex/379024.png)
,
![](http://thumb.1010pic.com/pic5/latex/379025.png)
.
(Ⅱ)由已知可知
![](http://thumb.1010pic.com/pic5/latex/78440.png)
,故
![](http://thumb.1010pic.com/pic5/latex/78441.png)
.
∵a
n+1≠0,∴a
n+2-a
n=2(n∈N
*).
于是 数列{a
2m-1}是以a
1=1为首项,2为公差的等差数列,∴a
2m-1=1+2(m-1)=2m-1,
数列{a
2m}是以a
2=2为首项,2为公差的等差数列,∴a
2m=2+2(m-1)=2m,
∴a
n=n(n∈N
*).
(Ⅲ)可知
![](http://thumb.1010pic.com/pic5/latex/379026.png)
.下面给出证明:
要比较T
n与
![](http://thumb.1010pic.com/pic5/latex/180754.png)
的大小,只需比较2T
n与log
2(2a
n+1)的大小.
由
![](http://thumb.1010pic.com/pic5/latex/78438.png)
,得
![](http://thumb.1010pic.com/pic5/latex/83633.png)
,
![](http://thumb.1010pic.com/pic5/latex/83634.png)
,
故
![](http://thumb.1010pic.com/pic5/latex/78446.png)
.
从而
![](http://thumb.1010pic.com/pic5/latex/78447.png)
.
![](http://thumb.1010pic.com/pic5/latex/379027.png)
=
![](http://thumb.1010pic.com/pic5/latex/379028.png)
因此2T
n-log
2(2a
n+1)=
![](http://thumb.1010pic.com/pic5/latex/379028.png)
-log
2(2n+1)
=
![](http://thumb.1010pic.com/pic5/latex/78450.png)
=
![](http://thumb.1010pic.com/pic5/latex/78451.png)
.
设
![](http://thumb.1010pic.com/pic5/latex/78452.png)
,
则
![](http://thumb.1010pic.com/pic5/latex/379029.png)
,
故
![](http://thumb.1010pic.com/pic5/latex/83642.png)
=
![](http://thumb.1010pic.com/pic5/latex/83643.png)
,
又f(n)>0,∴f(n+1)>f(n).
所以对于任意 n∈N
*都有
![](http://thumb.1010pic.com/pic5/latex/78454.png)
,
从而2T
n-log
2(2a
n+1)=log
2f(n)>0.
所以
![](http://thumb.1010pic.com/pic5/latex/379030.png)
.
即
![](http://thumb.1010pic.com/pic5/latex/379026.png)
.
分析:(I)利用
![](http://thumb.1010pic.com/pic5/latex/180753.png)
,其中a
1=1,a
n≠0,令n分别取1,2即可得出;
(II)由已知可知
![](http://thumb.1010pic.com/pic5/latex/78440.png)
,可得
![](http://thumb.1010pic.com/pic5/latex/78441.png)
.由于a
n+1≠0,转化为一个分奇数项和偶数项分别成等差数列:a
n+2-a
n=2
(n∈N
*). 即可得出通项a
n.
(III) 要比较T
n与
![](http://thumb.1010pic.com/pic5/latex/180754.png)
的大小,只需比较2T
n与log
2(2a
n+1)的大小.利用(II)和已知条件即可得出2T
n,令f(n)=2T
n-log
2(2a
n+1),比较f(n+1)与f(n)的大小即可得出结论.
点评:本题考查了数列的通项a
n与S
n之间的关系,分类讨论的思想方法,等差数列的通项公式,对数的运算性质,作差法和作商比较两个数的大小等知识与方法,熟练掌握它们是解题的关键.本题需要较强的计算能力和转化能力.