放缩法技巧全总结（尖子生解决高考数学最后一题之瓶颈之精华）

高考数学备考之放缩技巧

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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

n?4kk?122的值; (2)求证:

?1??kk?1n12?5. 3解析:(1)因为 (2)因为124n?112211,所以n212n ???1???2(2n?1)(2n?1)2n?12n?12n?12n?1k?14k?14n1111?25 1?,所以?1?1?2??1??????1????2??2?2???2k352n?12n?133??k?114n?1n?2n?12n?1?n2?4奇巧积累:(1)1?4?22n4n211 1? (2)1?1????2???124n?1?2n?12n?1?Cn?1Cn(n?1)n(n?1)n(n?1)n(n?1)42 (3)Tr?1r?Cn?1n!11111??r????(r?2) rr!(n?r)!nr!r(r?1)r?1rn (4)(1?1)n?1?1?1?n2?1115???? 3?2n(n?1)21?n?2?n n?2 (5)

111 (6)

?n?nnn2(2?1)2?1211 (7)2(n?1?n)?1?2(n?n?1) (8) ?2?1??1? ???nn?1n(2n?1)?2(2n?3)?2n?2n?12n?3?2 (9)

111?111?11??

????,????k(n?1?k)?n?1?kk?n?1n(n?1?k)k?1?nn?1?k?n11 (11)??1(n?1)!n!(n?1)!?2(2n?1?2n?1)?n222n?1?2n?1?n?211?n?22 (10) (11) (12)

2n2n2n2n?111

????n?1?n(n?2)n2nnnnnn?1(2?1)(2?1)(2?1)(2?1)(2?2)(2?1)(2?1)2?12?11n3?1n?n2???1111 ???????n(n?1)(n?1)?n(n?1)n(n?1)?n?1?n?11?n?1?n?1?1??????n?1?2n?n?111

?n?1n?1n (13) 2n?1?2?2n?(3?1)?2n?3?3(2n?1)?2n?2n?1?2?312n

?2n?13 (14)

k?211 (15)

??k!?(k?1)!?(k?2)!(k?1)!(k?2)!1?n?n?1(n?2)

n(n?1)i?j (15)

i2?1?j2?1i2?j2?i?j(i?j)(i2?1?j?1)2?i?1?2j?12?1

171例2.(1)求证:1?1?1?????(n?2) 22262(2n?1)35(2n?1) (2)求证:1?1?1???1?1?1 2416364n24n (3)求证:1?1?3?1?3?5???1?3?5???(2n?1)?22?42?4?62?4?6???2n2n?1?1

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(4) 求证：2(n?1?1)?1?1?1???1?2(2n?1?1)

23n

解析:(1)因为

111?11?,所以 ?????(2n?1)2(2n?1)(2n?1)2?2n?12n?1??(2i?1)i?1n12111111 ?1?(?)?1?(?)232n?1232n?1 (2)1?1?1???1?1(1?1???1)?1(1?1?1)

416364n4n2422n2 (3)先运用分式放缩法证明出1?3?5???(2n?1)?2?4?6???2n2n?1?n12n?1,再结合

1n?2?n?2?n进行裂项,最后就可以得到答案

(4)首先1n?2(n?1?n)?,所以容易经过裂项得到

1

n2(n?1?1)?1?12?13???再证

1n?2(2n?1?2n?1)?222n?1?2n?1?n?211?n?22而由均值不等式知道这是显然成立的，所以

1?12?13???1n?2(2n?1?1)

例3.求证:

6n1115?1?????2?

(n?1)(2n?1)49n31??1??2?2???214n?12n?12n?1?n2?n?414 解析:一方面:因为1,所以

?kk?1n1211?25 ?11?1?2????????1??2n?12n?1?33?3511n 另一方面:1?1?1???1?1?1?1????1??249n2?33?4n(n?1)n?1n?1 当n?3时,nn?1?6n1116n,当n?1时,?1?????2,

(n?1)(2n?1)49n(n?1)(2n?1)当n?2时,

6n111?1?????2,所以综上有

(n?1)(2n?1)49n6n1115?1?????2?

(n?1)(2n?1)49n3 例4.(2008年全国一卷) 设函数f(x)?x?xlnx.数列?an?满足0?a1?1.an?1?f(an).设b?(a1，1)，整数

k≥a1?b.证明:a?b.

k?1a1lnb 解析:由数学归纳法可以证明?an?是递增数列,故存在正整数m?k,使am?b,则

ak?1?ak?b,否则若am?b(m?k),则由0?a1?am?b?1知

amlnam?a1lnam?a1lnb?0,ak?1?ak?aklnak?a1??amlnam,因为?amlnam?k(a1lnb),

m?1m?1kk于是ak?1?a1?k|a1lnb|?a1?(b?a1)?b

例5.已知n,m?N?,x??1,Sm?1m?2m?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?nx

nm?1?nm?1?(n?1)m?1?(n?1)m?1?(n?2)m?1???1m?1?0?[km?1?(k?1)m?1]所以要证

?k?1n

nm?1?(m?1)Sn?(n?1)m?1?1只要证:

nn

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?[km?1?(k?1)m?1]?(m?1)?km?(n?1)m?1?1?(n?1)m?1?nm?1?nm?1?(n?1)m?1???2m?1?1m?1??[(k?1)m?1?km?1]k?1k?1k?1

故只要证

?[km?1?(k?1)m?1]?(m?1)?km??[(k?1)m?1?km?1],即等价于

k?1k?1k?1nnnkm?1?(k?1)m?1?(m?1)km?(k?1)m?1?km,即等价于1?m?1?(1?1)m?1,1?m?1?(1?1)m?1

kkkk而正是成立的,所以原命题成立. 例6.已知an?4n?2n,T?n解析:所以

2n,求证:T?T?T???T?3.

123n2a1?a2???anTn?41?42?43???4n?(21?22???2n)? 4(1?4n)2(1?2n)4n??(4?1)?2(1?2n)1?41?23

2n2n3?2n32nTn??n?1?n?1?n?1??4n44424?3?2n?1?222?(2n)2?3?2n?1(4?1)?2(1?2n)??2?2n?1??2n?133333?32n3?11? ???n?n?1?nn2(2?2?1)(2?1)2?2?12?1?2n 从而T?T?T???T?3?1?1?1?1???1?1??3

??123n2?3372n?12n?1?1?2例7.已知x1?1,x??n(n?2k?1,k?Z),求证:

?n?n?1(n?2k,k?Z)4 111?????2(n?1?1)(n?N*)4xxx2?x34x4?x52n2n?114证明: x2nx2n?1?14(2n?1)(2n?1)14?144n2?1?22n?144n2?212?n?2,因为 2n 2n?n?n?1,所以 所以

4x2nx2n?1?n?n?1?2(n?1?n)

111?????2(n?1?1)(n?N*)4xxx2?x34x4?x52n2n?1 二、函数放缩

例8.求证：ln2?ln3?ln4???ln3?3n?5n?6(n?N*).

2343n6 解析:先构造函数有lnx?x?1?lnx?1?1,从而ln2?ln3?ln4???ln3n?3n?1?(1?1???1)

nnxxn2343233因为111?11??111111?11? ?1????n????????????????n?n???n?2332?13??23??456789??2?3n?15?33??99?3n?1?5n ??????????????2?3n?1?3n???66?69??1827???所以ln2?ln3?ln4???ln3?3n?1?5n?3n?5n?6

234663n 例9.求证:(1)??2,ln2?ln3???lnn??????n23n2n2?n?1?(n?2)

2(n?1)4 / 24 解析:构造函数?lnn2lnx,得到lnn?2f(x)??xnn 2,再进行裂项lnn?1?1?1?1,求和后可以得到答案 n(n?1)n2n2?? 函数构造形式: lnx?x?1,lnn?n?1(??2) 例10.求证:1?1???23111?ln(n?1)?1???? n?12nn?11nn?1解析:提示:ln(n?1)?lnn?1?n???2?lnn?1?lnn???ln2 n函数构造形式: lnx?x,lnx?1?1x 当然本题的证明还可以运用积分放缩 如图,取函数f(x)?1, xy首先:SABCF11,从而,1 ???i???lnx|nn?i?lnn?ln(n?i)xnxn?in?inOnnEFAn-inDCBx取i?1有,1?lnn?ln(n?1), 所以有1?ln2,1?ln3?ln2,…,1?lnn?ln(n?1),1?ln(n?1)?lnn,相加后可以得到: 3n?1n2111?????ln(n?1) 23n?1另一方面SABDE11,从而有1 ?i???lnx|n??n?i?lnn?ln(n?i)n?ixn?in?ixnn取i?1有,1?lnn?ln(n?1), n?1所以有ln(n?1)?1?1???1,所以综上有1?1???1?ln(n?1)?1?1???1 2n23n?12n例11.求证:(1?1)(1?1)???(1?1)?e和1(1?)(1?2!3!n!9. 11)???(1?2n)?e813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:

,叠加之后就可以得到答案 3n(n?1)?131?ln(1?x)3(x?0)??(x?0)x?1xx?1(加强命题)

ln[n(n?1)?1]?2? 函数构造形式:

ln(x?1)?2? 例13.证明:ln23? ln3ln4lnnn(n?1)?????(n?N*,n?1)45n?14 解析:构造函数f(x)?ln(x?1)?(x?1)?1(x?1),求导,可以得到:

f'(x)?12?x,令?1?x?1x?1f'(x)?0有1?x?2,令f'(x)?0有x?2,

所以f(x)?f(2)?0,所以ln(x?1)?x?2,令x?n2?1有,lnn2?n2?1 所以lnnn?1?n?1,所以ln232?ln3ln4lnnn(n?1) ?????(n?N*,n?1)45n?14 例14. 已知a?1,a?(1?1)a?1.证明an?e2.

1n?1n2nn?n2

解析:

an?1?(1?

, 1111)an?n?(1??n)ann(n?1)n(n?1)22lnan?1?ln(1?5 / 24

然后两边取自然对数,可以得到

11?n)?lnann(n?1)2然后运用ln(1?x)?x和裂项可以得到答案) 放缩思路：

an?1?(1?1111lna?ln(1??)?lnan? ?)a?n?12nn2nn?n2n?n2?lnan?1111?n。于是lnan?1?lnan?2?n， n?n2n?n22n?1i?1?i?1n?1(lnai?1?lnai)??111(2?i)?lnan?lna1?1??ni?i211?()n?1 112?2??n?2.1n21?2即lnan?lna1?2?an?e2.

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：

an?1?(1?11)an??an?1?1?(1?1)(an?1)?

n(n?1)n(n?1)n(n?1)n?1n?1 ， 1111?ln(an?1)?ln(a2?1)?1??1)?.??[ln(ai?1?1)?ln(ai?1)]??i(i?1)nn(n?1)n(n?1)i?2i?2ln(an?1?1)?ln(an?1)?ln(1?即ln(an?1)?1?ln3?an?3e?1?e2.

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)?f(x)在(0,??)上是增函数；

x (II)当x1?0,x2?0时,证明:f(x1)?f(x2)?f(x1?x2)； (III)已知不等式ln(1?x)?x在x??1且x?0时恒成立， 求证：122ln22?111nln32?2ln42???ln(n?1)2?222(n?1)(n?2)34(n?1)(n?N*). 解析:(I)g'(x)?f'(x)x?f(x)?0,所以函数g(x)?f(x)在(0,??)上是增函数

x2x (II)因为g(x)?f(x)在(0,??)上是增函数,所以

x f(x1)x1?f(x1?x2)x1

?f(x1)??f(x1?x2)x1?x2x1?x2f(x2)f(x1?x2)x2 ??f(x2)??f(x1?x2)x2x1?x2x1?x2 两式相加后可以得到f(x1)?f(x2)?f(x1?x2) (3) f(x1)x1?f(x1?x2???xn)x1

?f(x1)??f(x1?x2???xn)x1?x2???xnx1?x2???xnf(x2)f(x1?x2???xn)x2……

??f(x2)??f(x1?x2???xn)x2x1?x2???xnx1?x2???xnf(xn)f(x1?x2???xn)xn

??f(xn)??f(x1?x2???xn)xnx1?x2???xnx1?x2???xn相加后可以得到:

f(x1)?f(x2)???f(xn)?f(x1?x2???xn)

所以x1lnx1?x2lnx2?x3lnx3???xnlnxn?(x1?x2???xn)ln(x1?x2???xn) 令

xn? ?11112222????22ln2?32ln3?42ln4???(n?1)2ln(n?1)?????1,有

2(1?n)

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