【答案】
分析:(1)將b
n=a
n-1代入2a
n=1+a
na
n+1,可得b
n的遞推關(guān)系式,整理變形可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/0.png)
,由等差數(shù)列的定義可得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/1.png)
為等差數(shù)列,故可求其通項(xiàng)公式,進(jìn)而求出b
n.
(2)結(jié)合(1)中的結(jié)論,寫出T
n+1-T
n的表達(dá)式,利用放縮法證明該差大于0即可.
(3)利用疊加法把
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/2.png)
轉(zhuǎn)化為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/3.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/4.png)
+…+T
2+T
1+S
1的形式,再結(jié)合(2)中的結(jié)論,利用T
n的單調(diào)性證明不等式.
解答:解:(1)由b
n=a
n-1,得a
n=b
n+1,代入2a
n=1+a
na
n+1,
得2(b
n+1)=1+(b
n+1)(b
n+1+1),
∴b
nb
n+1+b
n+1-b
n=0,從而有
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/5.png)
,
∵b
1=a
1-1=2-1=1,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/6.png)
是首項(xiàng)為1,公差為1的等差數(shù)列,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/7.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/8.png)
;(5分)
(2)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/9.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/11.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/12.png)
,
∴T
n+1>T
n;(10分)
(3)∵n≥2,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/14.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/15.png)
+…+T
2+T
1+S
1.
由(2)知
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/16.png)
≥
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/17.png)
≥…≥T
2≥T
1≥S
1,
∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/18.png)
,
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/19.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/20.png)
+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/21.png)
+…+T
2+T
1+S
1≥(n-1)T
2+T
1+S
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/22.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181038466242217/SYS201310241810384662422019_DA/23.png)
.(16分)
點(diǎn)評(píng):本題考查了數(shù)列和不等式的綜合應(yīng)用,應(yīng)用了構(gòu)造法、放縮法、疊加法等數(shù)學(xué)思想方法,難度較大.