【答案】
分析:(1)由曲線C:y=x
3,求導(dǎo)得切線斜率,切點Q
n的坐標(biāo)(a
n,a
n3),得切線方程,切線過點P
n-1(a
n-1,0),代入方程,得關(guān)于數(shù)列{a
n}項的關(guān)系式,變形得出數(shù)列{a
n}為等差數(shù)列,可求數(shù)列{a
n}的通項公式;
(2)把每一項的分子用錯位相減法都化為1,然后用等比數(shù)列的前n項和求解.
(3)法1,把
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/0.png)
分解為1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/1.png)
后用二項式定理,取前兩項即可;
法2,用數(shù)學(xué)歸納法:第一步,當(dāng)n=2時,結(jié)論成立;第二步,假設(shè)n=k時,結(jié)論成立,證明n=k+1時結(jié)論也成立.
解答:解:(1)∵y=x
3,∴y′=3x
2,設(shè)Q
n的坐標(biāo)為(a
n,a
n3),
則切線方程為y-a
n3=3a
n2(x-a
n),
切點為Q
1時,過點P
(1,0),
即:0-a
13=3a
12(1-a
1),
依題意a
1>0.所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/2.png)
.(2分)
當(dāng)n>1時,切線過點P
n-1(a
n-1,0),
即:0-a
n3=3a
n2(a
n-1-a
n),
依題意a
n>0,所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/3.png)
.(3分)
所以數(shù)列a
n是首項為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/4.png)
,
公比為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/5.png)
的等比數(shù)列.所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/6.png)
.(4分)
(2)記S
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/7.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/8.png)
,
因為
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/9.png)
,
所以
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/10.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/11.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/12.png)
.(5分)
兩式相減得:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/14.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/16.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/19.png)
.(7分)
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/20.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/22.png)
.(9分)
(3)①證法1:
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/24.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/25.png)
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/26.png)
.(14分)
②證法2:當(dāng)n=2時,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/27.png)
.(10分)
假設(shè)n=k時,結(jié)論成立,即
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/28.png)
,
則
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/29.png)
.
即n=k+1時.
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/30.png)
.(13分)
綜上,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131024181132002089463/SYS201310241811320020894018_DA/31.png)
,(n≥2,n∈N
*).(14分)
點評:本小題主要考查數(shù)列、導(dǎo)數(shù)、不等式和數(shù)學(xué)歸納法等知識,考查化歸與轉(zhuǎn)化的數(shù)學(xué)思想方法,以及邏輯推理,抽象概括能力,運算求解能力和創(chuàng)新意識,此題有點難度,需要同學(xué)們掌握.用錯位相減法求數(shù)列的前n項和,用時要觀察項的特征,是否是等差數(shù)列的項與等比數(shù)列的項的乘積.