2010高考数学备考之放缩技巧
放缩法精选大全
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???k352n?12n?1332??k?114n?1n?2n?12n?1?n2?4奇巧积累:(1)1?4?22n4n2111? (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)2n?2?n
(5)
111 (6) 1???n?22n(2n?1)2n?12n11 (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)
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?1?1n?n2???1111 ???????n(n?1)(n?1)?n(n?1)n(n?1)?n?1?n?1 (12) 1n3 ?11?n?1?n?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)
1k?211 (15) ???k!?(k?1)!?(k?2)!(k?1)!(k?2)!n(n?1)n?n?1(n?2)
22i2?j2 (15) i?1?j?1?i?j(i?j)(i2?1?j?1)2?i?ji?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???2nn2n?1?1
(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???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??49n22?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
a1lnb明:ak?1?b.
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:
?[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] 故
k?1k?1k?1nnn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?n2n,求证:T?T?T???T?3.
123n2a1?a2???annn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??n?1n24n44424?3?2?222?(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?11111?3 从而T?T?T???T?3??n?1?1??????n??123n2?3372?12?1?2例7.已知x1?1,x??n(n?2k?1,k?Z),求证:
?n?n?1(n?2k,k?Z)14x2?x31?14x4?x5???114x2nx2n?1??2(n?1?1)(n?N*)114
2,因为 2n证明:
4x2nx2n?1?4(2n?1)(2n?1)1444n2?1?22n?4n2?212?n? 2n?n?n?1,所以 所以
4x2nx2n?1?n?n?1?2(n?1?n)
1x2?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*).
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112?3???1?11??111111?11? ?1????????????????n?n???n?n32?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
n
放缩法技巧全总结(高考精品,吐血推荐,不看后悔一辈子)
