【答案】
分析:(1)利用平方差公式對題設(shè)中的等式化簡整理求得
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/0.png)
,進(jìn)而根據(jù)等差數(shù)列的定義判斷出數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/1.png)
是一個(gè)首項(xiàng)為1公差為1的等差數(shù)列.進(jìn)而根據(jù)首項(xiàng)和公差求得數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/2.png)
的通項(xiàng)公式,進(jìn)而根據(jù)a
n=S
n-S
n-1求得a
n.
(2)把(1)中的a
n代入b
n,進(jìn)而根據(jù)裂項(xiàng)法求得前n項(xiàng)的和,求得T
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/3.png)
,進(jìn)而利用
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/4.png)
推斷出
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/5.png)
,原式得證.
解答:解:(1)∵
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/6.png)
,(n≥2)
又b
n≥o,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/7.png)
,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/8.png)
,
又
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/9.png)
,所以數(shù)列
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/10.png)
是一個(gè)首項(xiàng)為1公差為1的等差數(shù)列.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/11.png)
,s
n=n
2.
當(dāng)n≥2,a
n=S
n-S
n-1=n
2-(n-1)
2=2n-1;a
1=1適合上式,∴a
n=2n-1(n∈N).
(2)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/12.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/13.png)
,
T
n=b
1+b
2++b
n![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/14.png)
;
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/16.png)
∵n∈N,∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/17.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/18.png)
,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/19.png)
,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212936645800480/SYS201310232129366458004020_DA/20.png)
.
點(diǎn)評:本題主要考查了等差關(guān)系的確定和數(shù)列的求和,數(shù)列和不等式的綜合運(yùn)用.作為高考的必考內(nèi)容,數(shù)列題常與不等式,函數(shù)等問題綜合考查,綜合性較強(qiáng).