放缩技巧
（高考数学备考资料）
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?124n2?11?n2n的值; (2)求证:?1?5.
2k?1k3解析:(1)因为
?211,所以n212n ???1???2(2n?1)(2n?1)2n?12n?12n?12n?1k?14k?14 (2)因为
n1111?25 1?,所以?1?1?2??1??????1????2?2?2???2n?12n?1?33?35k?1k14n?1?2n?12n?1?n2?41奇巧积累
:
(1)
1441? ?1?2?2?2???2n4n4n?1?2n?12n?1?r?1r?Cn? (2)
1211 ???2CCn(n?1)n(n?1)n(n?1)n(n?1)1n?1
(3)T1n!11111??r????(r?2) rr!(n?r)!nr!r(r?1)r?1rn (4)(1?1)n?1?1?1?1???n2?13?215?
n(n?1)21?n?2?n n?2?2n?12n?3?211?n?1(2n?1)?2(2n?3)?2n (5)
111?n?nnn2(2?1)2?12 (6)
21?1 (7)2(n?1?n)?1?2(n?n?1) (8) ?????n?
n (9)
111?111?11?? ????,????k(n?1?k)?n?1?kk?n?1n(n?1?k)k?1?nn?1?k?n11 ??(n?1)!n!(n?1)! (10) (11)
1n?2(2n?1?2n?1)?222n?1?2n?1?n?211?n?22
(11) (12) (13) (14)
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?1
1?n?1?n?1?1??????n?1?2n?n?12n?1nnn11
?n?1n?1nn
1?n?n?1(n?2) n(n?1)2n12n?2?2?(3?1)?2?3?3(2?1)?2?2?1??n?32?13k?211 ??k!?(k?1)!?(k?2)!(k?1)!(k?2)! (15)
22 (15) i?1?j?1?i2?j2(i?j)(i?1?2i?jj?1)2?i?ji?1?2j?12?1
例2.(1)求证:1?11171?2?????(n?2) 2262(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)?2n?1?1
22?42?4?62?4?6???2nn(4) 求证：2(n?1?1)?1?1?1???1?2(2n?1?1)
23解析:(1)因为
111?11?,所以 ?????2(2n?1)(2n?1)2?2n?12n?1?(2n?1)?(2i?1)i?1n12111111 ?1?(?)?1?(?)232n?1232n?1 (2)1?1?1???1?1(1?1???1)?1(1?1?1)
222416364n42n4n (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
1n?2?n?2?n进行裂项,最后就可以得到答案
(4)首先再证
1n1n?2(n?1?n)?2n?1?n22,所以容易经过裂项得到2(n?1?1)?1?1?1???1
23n而由均值不等式知道这是显然成立的，
?2(2n?1?2n?1)?2n?1?2n?1?n?211?n?22所以1?1?1???1?2(2n?1?1)
23n
例3.求证:
6n1115?1?????2?
(n?1)(2n?1)49n31?n21??1?2?2???14n?12n?12n?1?2?n?414解析: 一方面: 因为,所以
?kk?1n1211?25 ?11?1?2????????1??2n?12n?1?33?35 另一方面: 1?1?1???1?1?1?1???249n2?33?411n
?1??n(n?1)n?1n?1 当n?3时,
当n?2时,所以综上有
6n111n6n,当n?1时,?1?????2?(n?1)(2n?1)49nn?1(n?1)(2n?1),
6n111?1?????2,
(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?a明:ak?1?b.
11)，整数k≥a1?b?1.an?1?f(an).设b?(a1，.证
a1lnb解析: 由数学归纳法可以证明?an?是递增数列, 故 若存在正整数m?k, 使am?b, 则ak?1?ak?b,
若am?b(m?k),则由0?a1?am?b?1知amlnam?a1lnam?a1lnb?0,a?a?alna?a?kalna,
?mmk?1kkk1m?1因为
?am?1kmlnam?k(a1lnb),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?n[km?1?(k?1)m?1]所以要证
?k?1 nm?1?(m?1)Sn?(n?1)m?1?1只要证:
?[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?1nnn
故只要证
?[kk?1nm?1?(k?1)m?1]?(m?1)?k??[(k?1)m?1?km?1],
mk?1k?1nn即等价于km?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 而正是成立的,所以原命题成立.
kkkk
例6.已知an?4n?2n,T?n2na1?a2???an,求证:T?T?T???T?3.
123n2nn解析:T?41?42?43???4n?(21?22???2n)?4(1?4)?2(1?2)?4(4n?1)?2(1?2n)
n1?41?23所以
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?1333332n
?32n3?11? ???n?n?1?nn2(2?2?1)(2?1)2?2?12?1?1111 从而T?T?T???T?3???1??????n123n2?3372?112n?1?3
???1?2
例7.已知x1?1,x??n(n?2k?1,k?Z),求证:
?n?n?1(n?2k,k?Z)1414x2?x3??114x4?x5????214x2nx2n?1?2(n?1?1)(n?N*)
证明:
41x2nx2n?1?(2n?1)(2n?1)?144n?112?14,
4n222?n22n?2(n?1?n)因为
2n?n?n?1,所以
4x2nx2n?1?2n?n?n?1所以
41x2?x3?14x4?x5???14x2nx2n?1?2(n?1?1)(n?N*)
二、函数放缩
例8.求证：ln2?ln3?ln4???ln3?3n?5n?6(n?N*).
n23436n 解析:先构造函数有lnx?x?1?lnx?1?1,从而ln2?ln3?ln4??xx234ln3n111 ?n?3n?1?(????n)2333n?1n?1cause1?1??23?39?31?11??111111?11?5?33??9?1???????n?n????????????????n?n???n????????n?1?32?13?6?69??1827?3?23??456789??2?2?3?5n
???6?所以ln2?ln3?ln4???ln3?3n?1?5n?3n?5n?6
n234366n
???2 例9.求证:(1)??2,ln2?ln3???lnn?2n?n?1(n?2) ???23n2(n?1) 解析:构造函数?lnn2lnx,得到lnn?2f(x)??nnx2,再进行裂项lnn?1?1?1?22nn1,求和后可以得到答案 n(n?1) 函数构造形式: lnx?x?1,lnn??n??1(??2)
例10.求证:1?1???1?ln(n?1)?1?1???1 23n?12n解析:提示:ln(n?1)?lnn?1?n???2?lnn?1?lnn???ln2 nn?11 x1nn?1函数构造形式: lnx?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, xEFOAn-inDCBx首先:SABCF1??xn?innn,从而,1?i?1?lnx|n?lnn?ln(n?i) n?i?nn?ix取i?1有,1?lnn?ln(n?1), 所以有1?ln22,1?ln3?ln2,…,1?lnn?ln(n?1),3n1?ln(n?1)?lnn,相加后可以得到: n?1111?????ln(n?1) 23n?1另一方面S取i?1有,ABDE1??n?ixn,从而有1?i?n?i1 ?lnx|nn?i?lnn?ln(n?i)?n?ixn1?lnn?ln(n?1), n?1?111 ?ln(n?1)?1????n?12n所以有ln(n?1)?1?1???1,所以综上有1?1??2n23
例11.求证:(1?1)(1?1)???(1?1)?e和(1?1)(1?2!3!n!9.解析:构造函数后即可证明 11)???(1?2n)?e813,叠加之后就可以得到答案 3n(n?1)?1
例12.求证:(1?1?2)?(1?2?3)???[1?n(n?1)]?e2n?3 解析:
函数构造形式:
例13.证明:ln2?ln3?ln4???34531?ln(1?x)3(x?0)??(x?0)x?1xx?1ln[n(n?1)?1]?2?ln(x?1)?2?(加强命题) lnnn(n?1)?(n?N*,n?1)n?14f'(x)?0有1?x?2,令f'(x)?0有x?2,
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?1 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?12,所以ln2?ln3?ln4???345lnnn(n?1) ?(n?N*,n?1)n?14
例14. 已知a1?1,an?1?(1?11证明a?e2. n)an?n.n?n22 解析:
an?1?(1?, 1111)an?n?(1??n)ann(n?1)n(n?1)22n?1然后两边取自然对数,可以得到lna?ln(1? 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：1111an?1?(1?n?n2?2n)an?lnan?1?ln(1?n?n2?2n)?lnan??lnan?11?nn?n22n?1i?1。于是
lnan?1?lnan?11?nn?n22，
?i?1n?1(lnai?1?lnai)??111(2?i)?lnan?lna1?1??ni?i211?()n?1112?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：
111a?(1?)a??n?1n(n?1)nn(n?1)an?1?1?(1?n(n?1))(an?1)?ln(an?1?1)?ln(an?1)?ln(1?n?1 n?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?2即ln(an?1)?1?ln3?an?3e?1?e2.
例16.(2008年福州市质检)已知函数f(x)?xlnx.若a?0,b?0,证明:f(a)?(a?b)ln2?f(a?b)?f(b). 解析:设函数g(x)?f(x)?f(k?x),(k?0)
Qf(x)?xlnx,?g(x)?xlnx?(k?x)ln(k?x),
?0?x?k.Qg?(x)?lnx?1?ln(k?x)?1?ln令g?(x)?0,则有x ,k?xx2x?kk?1??0??x?k.k?xk?x2kkk∴函数g(x)在[,k）上单调递增，在(0,k]上单调递减.∴g(x)的最小值为g()，即总有g(x)?g().
2222
而g(k)?f(k)?f(k?k)?klnk?k(lnk?ln2)?f(k)?kln2,
2222?g(x)?f(k)?kln2,
即f(x)?f(k?x)?f(k)?kln2. 令x?a,k?x?b,则k?a?b.
?f(a)?f(b)?f(a?b)?(a?b)ln2. ?f(a)?(a?b)ln2?f(a?b)?f(b).
高考数学_压轴题_放缩法技巧全总结(最强大)
