2010高考数学备考之放缩技巧
证明数列型不等式，因其思维跨度大、构造性强，需要有较高的放缩技巧而充满思考性和挑战性，能全面而综合地考查学生的潜能与后继学习能力，因而成为高考压轴题及各级各类竞赛试题命题的极好素材。这类问题的求解策略往往是：通过多角度观察所给数列通项的结构，深入剖析其特征，抓住其规律进行恰当地放缩；其放缩技巧主要有以下几种： 一、裂项放缩 例1.(1)求?k?1n24k2的值; (2)求证:
?1?2(2n?1)(2n?1)?n?k?11k2?53. ,所以
n解析:(1)因为 (2)因为
24n?1212n?1?12n?1?k?124k2?1?1?12n?1?2n2n?1
1n2n111?25?111?,所以?1?1?2?????????1??2???2???22n?12n?1?33?35k?1k14n?12?2n?12n?1?n?414奇巧积累:(1) (3)T1n2?44n2?1? ?1?2???4n?1?2n?12n?1?42 (2)
C11n?1C2n?2(n?1)n(n?1)?1n(n?1)?1n(n?1)
r?1?Cn?r1nr?n!r!(n?r)!n?1r?1r!?1r(r?1)?1r?1?1r(r?2)
(4)(1? (5)
n1n1)?1?1?12?1nn12?1?12n?13?2???1n(n?1)?521
?n?2?n2(2?1)n? (6)
?2(n?n?1)
n?2 (7)2( (9)
n?1?1n)?1n (8)
1?111?2?????n?n?1n(2n?1)?2(2n?3)?2?2n?12n?3?211?111?11?? ????,????k(n?1?k)?n?1?kk?n?1n(n?1?k)k?1?nn?1?k?n(n?1)!?1n!?1(n?1)!
n?12 (10) (11)
1n?2(2n?1?2n?1)?222n?1?2n?1?n?212? (11)
n2n2(2?1)?2nnn(2?1)(2?1)?2nnn(2?1)(2?2)?2nn?1n?1(2?1)(2?1)?12n?1?1?12?1n(n?2)
(12) (13) (14)
1n3?1n?n2?????n(n?1)(n?1)?11n(n?1)????n(n?1)??11n?1?n?1
????1n?1????n?1?1n?1?2nnn?1?1n?1n?1n?1n
2n2n?1?2?2?(3?1)?2?3?3(2?1)?2?2?1?k?21(k?1)!1(k?2)!nn3?12?1n?2n
n?n?1(n?2)
3k!?(k?1)!?(k?2)!22?? (15)
1n(n?1)?
(15)
i?1?i?jj?1?i?j2222?i?ji?1?2(i?j)(i?1?j?1)j?12?1
例2.(1)求证:1?132?152???1(2n?1)2?76?12(2n?1)(n?2)
(2)求证:14?116?136????14n2?12?14n
?2n?1?1
(3)求证:12?1?32?41?3?52?4?6???121?3?5???(2n?1)2?4?6???2n13???1n? (4) 求证：2(解析:(1)因为 (2)141161(2n?1)2n?1?1)?1??2(2n?1?1)
111111(?)?1?(?)232n?1232n?1?1(2n?1)(2n?1)?1?11????2?2n?12n?1?,所以
1n2n?(2i?1)i?112?1?
??136???14n2?14(1?122???)?14(1?1?1n)
1n?2 (3)先运用分式放缩法证明出1?3?5???(2n?1)?2?4?6???2n12n?1,再结合
?n?2?n进行裂
项,最后就可以得到答案 (4)首先
1n?2(n?1?n)?2n?1?n,所以容易经过裂项得到
1n2(n?1?1)?1?12?13???
而由均值不等式知道这是显然
再证
1n?2(2n?1?2n?1)?222n?1?2n?1?n?212?n?12成立的，所以1?例3.求证:
6n12?13???14191n?1n22(2n?1?1) 53(n?1)(2n?1)?1??????
,所以
解析:一方面:因为
1n2?21n?14?1??1?2???24n?1?2n?12n?1?4n?kk?11211?25?11?1?2?????????1?2n?12n?1?33?35
1n(n?1)1n?119nn?11n2 另一方面:1? 当n?3时,
当n?2时,
n14?19???1n2?1?12?3?13?4???6n?1?14? ,
n?1?6n(n?1)(2n?1),当n?1?14?19?1时,
(n?1)(2n?1)???1n2?1?????6n(n?1)(2n?1),所以综上有
6n(n?1)(2n?1)?1?14?19???1n2?53
.
.设
例4.(2008年全国一卷) 设函数f(x)?x?xlnx.数列?an?满足0?ab?(a1，1)1?1an?1?f(an)，整数k≥a1?ba1lnb.证明:ak?1?b.
?k 解析:由数学归纳法可以证明?an?是递增数列,故存在正整数mak?1?ak?b,使am?b,则
,否则若a
m?b(m?k),则由0?a1?am?b?1知
amlnam?a1lnam?a1lnb?0,a?ak?aklnak?a1?k?1k?am?1mlnam,因为
k?am?1mlnam?k(a1lnb),
于是ak?1?a1?k|a1lnb|?a1?(b?a1)?b
例5.已知n,m?N?,x??1,Sm?1?2nmm?3m???nm,求证:
nm?1?(m?1)Sn?(n?1)m?1?1.
解析:首先可以证明:(1?x)
n?1?nxm?1
nm?1nm?1?nm?1?(n?1)m?1?(n?1)?(n?2)???1m?1?0??[kk?1m?1?(k?1)m?1]所以要证
nm?1?(m?1)Sn?(n?1)nm?1?1只要证:
nm?1?[kk?1m?1?(k?1)m?1]?(m?1)?kk?1nm?(n?1)m?1?1?(n?1)nnm?nm?1?nm?1?(n?1)m?1???2m?1?1m?1??[(k?1)k?1m?1?km?1] 故只要证
km?1?[kk?1m?1m?1?(k?1)m?1]?(m?1)?kk?1m?1m??[(k?1)k?1m?1?km?1],即等价于
1km?1?(k?1)?(m?1)km?(k?1)?k,即等价于1?m?1k?(1?),1?m?1k?(1?1k)m?1
而正是成立的,所以原命题成立. 例6.已知a解析:T所以
Tn?24332nnn?4?2nn,Tn?2na1?a2???ann12,求证:Tn1?T2?T3???Tn?nn32.
n?4?4?4???4?(2?2???2)?1234(1?4)1?4?2(1?2)1?2?43(4?1)?2(1?2)nn?n24n?1n(4?1)?2(1?2)32(2?2nnn?43?n?124n?1n?2?23?23??2n?13?24n?1nn?1?3?2?2??n2n22?(2)?3?2?132n
???1)(2?1)?3?11??n?1?n?2?2?12?1?
从而T1?T2?T3???Tn?3?111113??n?1?1??????n??2?3372?12?1?2例7.已知x141?n(n?2k?1,k?Z),求证: ?1,x??n?n?1(n?2k,k?Z)?14x2?x3x4?x5???14x2nx2n?1?2(n?1?1)(n?N*)
22n证明:
41x2nx2n?1n??41(2n?1)(2n?1)14?144n?1?22n?2?144n2?12?n?,因为
2n?n?1,所以
x2nx2n?114?2n?n?1?2(n?1?n) 所以
41x2?x3?14x4?x5???x2nx2n?12(n?1?1)(n?N*)
二、函数放缩
例8.求证：ln2?2ln33?ln44???ln331xnn?3?n5n?66(n?N).
*n 解析:先构造函数有lnx?x?1?因为12?13???13nlnxx?1?,从而ln2?ln3?ln4???ln32343n?3?1?(n12?13???13n)
?11???????23?11? ?111111??1???n????????????n?n2?13??456789??2
?33???????6?69?5n?1?3n?19?3?9???n??????n?13?1827??2?3?5n???6?n
5n?66所以ln2?2ln33?ln44???ln33?nn?3?1?n5n6?3??
(n?2)
1n2 例9.求证:(1)? 解析:构造函数以得到答案 函数构造形式: f(x)??2,ln22??ln33??????lnnn??2n?n?12(n?1)22lnxx,得到lnnn??lnnn22,再进行裂项lnnn2?1??1?1n(n?1),求和后可lnx?x?1,lnn??n??1(??2) 例10.求证:12?13???n?1n1n?1?nn?11x?ln(n?1)?1?21n?1n12???n1n ???ln2解析:提示:ln(n?1)?函数构造形式: ln????ln?ln n?1lnx?x,lnx?1? y当然本题的证明还可以运用积分放缩 如图,取函数f(x)?1, x首先:SnABCF??n?i1x,从而,1?i?nn?n?i1x?lnx|n?i?lnn?ln(n?i)n EFOAn-inDCBx取i?1有,1n?lnn?ln(n?1), ,…,1n?lnn?ln(n?1)所以有12?ln2,13?ln3?ln2,1n?1?ln(n?1)?lnn,相加后可以得到: 12?13???1n?1?ln(n?1)n 1n?in另一方面S取i?1有,ABDE??n?i1x,从而有?i??n?i1x?lnx|n?i?lnn?ln(n?i)n 1n?1?lnn?ln(n?1), 所以有ln(n?1)?1?1???1,所以综上有1?1???2n1n?118123?ln(n?1)?1?12???1n 例11.求证:(1?12!)(1?13!)???(1?1n!)?e和(1?1)(1?9)???(1?132n)?e.
解析:构造函数后即可证明
例12.求证:(1?1?2)?(1?2?3)???[1?n(n?1)]?e2n?3
解析:
ln[n(n?1)?1]?2?3n(n?1)?1,叠加之后就可以得到答案
(加强命题)
函数构造形式:ln(x?1)?2?3x?1(x?0)?1?ln(1?x)x?3x?1(x?0) 例13.证明:ln2?ln3?ln4???345lnnn?1?n(n?1)4(n?N*,n?1) 解析:构造函数
f(x)?f(x)?ln(x?1)?(x?1)?1(x?1),求导,可以得到:
2?xx?11x?1?1?,令f'(x)?0有1?x?2,令f'(x)?0有x,令xn(n?1)42?2,
22 所以
f(x)?f(2)?0,所以ln(x?1)?x?2n?12?n?1有,lnn?n?1
(n?N*,n?1) 所以lnnn?1?,所以ln2?ln3?ln4???345lnnn?1?
例14. 已知a 解析:
an?1?(1?11?1,an?1?(1?1n?n12)an?12n.证明a,
n?e2.
n(n?1))an?12n?(1?n(n?1)?12n)an然后两边取自然对数,可以得到
lnan?1?ln(1?1n(n?1)?12n)?lnan
然后运用ln(1?x)?x和裂项可以得到答案) 放缩思路：
an?1?(1?21n?n?12n)an?lnan?1?ln(1?1n?n2?12n)?lnan?
?lnan?n?11n?n2?12n。于是lna11n?1?lnan?1n?n12?12n，
n?1?i?1(lnai?1?lnai)??i?11n?11?()112(2?i)?lnan?lna1?1???2??n?2.1nn2i?i21?22即lnan?lna1?2?an?e.
注：题目所给条件ln(1?x)?x（x?0）为一有用结论，可以起到提醒思路与探索放缩方向的作用；当然，本题还可用结论2n?n(n?1)(n?2)来放缩：
111an?1?(1?n(n?1))an?n(n?1)?an?1?1?(1?n(n?1))(an?1)?ln(an?1?1)?ln(an?1)?ln(1?n?11n(n?1)n?1)?1n(n?1).
1n?1??[ln(ai?1?1)?ln(ai?1)]?i?2?i?21i(i?1)?ln(an?1)?ln(a2?1)?1?2，
即ln(an?1)?1?ln3?an?3e?1?e.
例15.(2008年厦门市质检) 已知函数f(x)是在(0,??)上处处可导的函数,若
x?f'(x)?f(x)在x?0上恒成立.
(I)求证：函数
g(x)?12f(x)x上是增函数；
在(0,??)1212 (II)当x?0,x?0时,证明:f(x)?f(x)?f(x?x)；
(III)已知不等式ln(1?x)?x在x??1且x?0时恒成立， 求证：
122ln2?2132ln3?2142ln4???21(n?1)2ln(n?1)?2n2(n?1)(n?2)(n?N).*
解析:(I)g'(x)? (II)因为
f(x)xf'(x)x?f(x)x2?0,所以函数
g(x)?f(x)x在(0,??)上是增函数
g(x)?在(0,??)上是增函数,所以