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?12的值; (2)求证:
?kk?1n12?5. 3解析:(1)因为
212n 211,所以n?1??????22n?12n?14n2?1(2n?1)(2n?1)2n?12n?1k?14k?1n111?25?111?,所以?1?1?2????????1?? ????2?2??2n?12n?1?33?35k?1k14n2?1n22?2n?12n?1?n?4 (2)因为114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?2) rrr!(n?r)!nr!r(r?1)r?1rn (4)(1?1)n?1?1?1?1?L?1?3
n2?13?2n(n?1)1111 (6) (5)?n?2?n ??nnnn2(2?1)2?12n?221?111 (7)2(n?1?n)?1?2(n?n?1) (8) ? ?????n?n?1n(2n?1)?2(2n?3)?2n?2n?12n?3?2111?111?11??
????,????k(n?1?k)?n?1?kk?n?1n(n?1?k)k?1?nn?1?k?n11 (11)1222 (10) ???2(2n?1?2n?1)??(n?1)!n!(n?1)!n2n?1?2n?111n??n?22 (9)
2n2n2n2n?111 (11) n?????(n?2)
(2?1)2(2n?1)(2n?1)(2n?1)(2n?2)(2n?1)(2n?1?1)2n?1?12n?1 (12)
1n3?1n?n2???1111 ???????n(n?1)(n?1)?n(n?1)n(n?1)?n?1?n?111?n?1?n?1 ???????n?1n?12n??11
?n?1n?1nnn (13) (14)
2n?12n12n?2?2?(3?1)?2?3?3(2?1)?2?2?1??n?
32?13nnk?211 (15) ??k!?(k?1)!?(k?2)!(k?1)!(k?2)!i2?j2(i?j)(i?1?21?n?n?1(n?2)
n(n?1)?1
22 (15) i?1?j?1?i?jj?1)2?i?ji?1?2j?12 例2.(1)求证:1?1?1???2235171??(n?2) 262(2n?1)(2n?1)(2)求证:
111111 ?????2??4163624n4n(3)求证:
11?31?3?51?3?5???(2n?1)??????2n?1?1 22?42?4?62?4?6???2n(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)
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1?2(n?1?n)?n2n?1?n22,所以容易经过裂项得到2(n?1?1)?1??n?211?n?2212?13???1n
再证1n?2(2n?1?2n?1)?而由均值不等式知道这是显然成立的，
2n?1?2n?1所以1?12?13???1n?2(2n?1?1)
例3.求证:
6n1115?1?????2?
(n?1)(2n?1)49n311?25?1111?,所以n1解析: 一方面: 因为1?1?4?2??1?2???????1?? ?????22n?12n?1?3314n2?1?2n?12n?1??35k?1kn22n?4 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)49n,
6n1115?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).
1)，整数k≥设b?(a1，a1?b.证明:ak?1?b. 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若am?b(m?k),则由0?k, 使am?b, 则ak?1?ak?b,
?a1?am?b?1知
kamlnam?a1lnam?a1lnb?0,ak?1?ak?aklnak?a1??amlnam,
m?1因为?amlnam?k(a1lnb),于是ak?1?a1?k|a1lnb|?a1?(b?a1)?b
m?1km?1?(m?1)Sn?(n?1)m?1?1. 例5.已知n,m?N?,x??1,Sm?1m?2m?3m???nm,求证: n 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 nnm?1?(m?1)Sn?(n?1)m?1?1只要证:
nn?[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
?[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.
kkkknn例6.已知an?4?2,Tn?32n,求证:T1?T2?T3???Tn?.
2a1?a2???an1?41?23nn解析:T?41?42?43???4n?(21?22???2n)?4(1?4)?2(1?2)?4(4n?1)?2(1?2n)
n所以
Tn?2n4n(4?1)?2(1?2n)3?2n2n3?2n32n????4n?144n?124n?1?3?2n?1?222?(2n)2?3?2n?1n?1n?1??2?2??23333
32n3?11?
????n?n?1?nn2(2?2?1)(2?1)2?2?12?1? 从而T1?T2?T3???Tn?3?11111?3?n?1?? ?1??????n2?3372?12?1?21x2?x3?14例7.已知x1?1,x??n(n?2k?1,k?Z),求证:
?n4?n?1(n?2k,k?Z)?14x4?x51????14x2nx2n?1?2(n?1?1)(n?N*)
证明:
41x2nx2n?1?14(2n?1)(2n?1)
4?1144n?1?224n2?222n2?nn?1?,
因为
2n?n?n?1,所以
x2nx2n?12n?n?2(n?1?n)所以
41x2?x3?14x4?x5???14x2nx2n?1?2(n?1?1)(n?N*)
二、函数放缩
ln2ln3ln4ln3n5n?6 例8.求证：?????n?3n?(n?N*).
23436n 解析:先构造函数有lnx?x?1?lnx?1?1,从而ln2?ln3?ln4???ln3?3n?1?(1?1???1)
nnxx2343233111?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
n234366nln2?ln3?lnn?2n2?n?1 例9.求证:(1)??2,????????(n?2)
2(n?1)23n?2lnn解析:构造函数f(x)?lnx,得到lnn?lnn,再进行裂项2nxn?n22?1?11,求和后可以得到答案 ?1?n(n?1)n2 函数构造形式:
例10.求证:lnx?x?1,lnn??n??1(??2) 11111?????ln(n?1)?1???? 23n?12n解析:提示:ln(n?1)?lnn?1n2n?1n?????ln?ln???ln2 nn?11nn?11 xy函数构造形式: lnx?x,lnx?1?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:SABCFn1,从而,11 ???i???lnx|nn?i?lnn?ln(n?i)xnxn?in?in取i?1有,1?lnn?ln(n?1), n2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: 1?1???1?ln(n?1) 3nn?123n?1另一方面S取i?1有,ABDE11?i???,从而有n?in?ixn1 ?lnx|nn?i?lnn?ln(n?i)?n?ixn1?lnn?ln(n?1), n?111111所以有ln(n?1)?1?1???1,所以综上有?????ln(n?1)?1????
23n?12n2n
例11.求证:(1?
例12.求证:(1?1?2)?(1?2?3)???[1?n(n?1)]?e2n?3 解析:ln[n(n?1)?1]?2?111111)(1?)???(1?)?e和(1?)(1?)???(1?2n)?e.解析:构造函数后即可证明 2!3!n!98133,叠加之后就可以得到
n(n?1)?1答案 函数构造形式:ln(x?1)?2?31?ln(1?x)3(x?0)??(x?0)(加强命题) x?1xx?1
例13.证明:
ln2ln3ln4lnnn(n?1)??????(n?N*,n?1) 345n?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?1?,令f'(x)?0有1?x?2,令f'(x)?0有x?2, x?1x?1 所以 所以
f(x)?f(2)?0,所以ln(x?1)?x?2,令x?n2?1有,lnn2?n2?1
ln2ln3ln4lnnn(n?1)lnnn?1,所以??????(n?N*,n?1) ?345n?14n?12211a?e例14. 已知a1?1,an?1?(1?2证明 )an?n.nn?n2解析: an?1?(1?1)an?1?(1?1?1)an,
n(n?1)n(n?1)2n2n然后两边取自然对数,可以得到ln然后运用ln(1?an?111?ln(1??)?lnan
n(n?1)2nx)?x和裂项可以得到答案)
1111lna?ln(1??)?lnan? ?n)an?n?12n2n?n2n?n2。于是lnan?1?lnan?放缩思路：an?1?(1?11?lnan?2?nn?n2?i?1n?111?，
n2?n2n(lnai?1?lnai)??i?1n?1an?lna1?2?a?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即ln11?()n?1 111112(2?i)?lnan?lna1?1???2??n?2.1nn2i?i21?22 n?n(n?1)(n?2)来放缩：
11)an??an?1?1?(1?1)(an?1)? n(n?1)n(n?1)n(n?1)11 )?.?n(n?1)n(n?1)n?1n?1an?1?(1?ln(an?1?1)?ln(an?1)?ln(1??[ln(ai?1?1)?ln(ai?1)]??i?2i?211 ?ln(an?1)?ln(a2?1)?1??1，
i(i?1)n即ln(an?1)?1?ln3?an?3e?1?e2.
例16.(2008年市质检)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?lnx, k?x令g?(x)?0,则有x2x?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().
2222kkkk 而g()?f()?f(k?)?kln?k(lnk?ln2)?f(k)?kln2,
2222
放缩法技巧全总结材料
