高考数学压轴题放缩法技巧全总结最强大

高考数学压轴题放缩法技巧全总结最强大

文档编制序号：[KKIDT-LLE0828-LLETD298-POI08]

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?1n24k?1222的值; (2)求证:??15?. 23k?1kn解析:(1)因为 (2)因为

4n?11211,所以n212n ???1???2(2n?1)(2n?1)2n?12n?12n?12n?1k?14k?141?n2n1111?25 1?,所以?1?1?2??1??????1????2?2?2???2n?12n?1?33?35k?1k14n?1?2n?12n?1?n2?4奇巧积累

:

(1)

2111441? (2)1?1 ????2?2?2???122n4n4n?1?2n?12n?1?Cn?1Cn(n?1)n(n?1)n(n?1)n(n?1) (3)Tr?1?Cnr?n1n!11111??r????(r?2) rr!(n?r)!nr!r(r?1)r?1rn (4)(1?1)n?1?1?1?1???2?13?215? n(n?1)21?n?2?n n?221?111????n?n?1(2n?1)?2(2n?3)?2n?2n?12n?3?2 (5)

111?n?nn2(2?1)2?12n (6)

(7)2( (9)

n?1?n)?1n?2(n?n?1) (8) ??

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?112n?n?32?13 (13) 2n?1?2?2n?(3?1)?2n?3?3(2n?1)?2n?2n?1?2n (14)

n?n?1(n?2)

k?211 (15) 1???k!?(k?1)!?(k?2)!(k?1)!(k?2)!n(n?1)

(15)

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

416364n24n (3)求证:1?1?3?1?3?5???1?3?5???(2n?1)?2n?1?1

22?42?4?62?4?6???2n???1n(4) 求证：2(n?1?1)?1?12?13?2(2n?1?1)

解析:(1)因为 (2)1?4111?11??????2(2n?1)(2n?1)2?2n?12n?1?(2n?1),所以

?(2i?1)i?1n12111111 ?1?(?)?1?(?)232n?1232n?111111111????2?(1?2???2)?(1?1?) 163644n4n2n12n?1 (3)先运用分式放缩法证明出1?3?5???(2n?1)?2?4?6???2n,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(2n?11?n?22n?1?1)?1?12?13???1n

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?所以1?12?13???1n?2(2n?1?1)

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???4911111n ?1??????1??2n2?33?4n(n?1)n?1n?1 当n?3时,

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,

当n?2时,

所以综上有

,

6n1115?1?????2?

(n?1)(2n?1)49n31)，整数?1.an?1?f(an).设b?(a1，例4.(2008年全国一卷)设函数f(x)?x?xlnx.数列?an?满足0?ak≥a1?b.证明:a?b. k?1a1lnb1

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,ak?1?ak?aklnakk因为?amlnam?k(a1lnb),于是ak?1?a1?k|a1lnb|?a1?(b?a1)?b

m?1?a1??amlnam,

km?1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

n 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?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?[km?1?(k?1)m?1]?(m?1)?km??[(k?1)m?1?km?1],

k?1k?1k?1nnn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,Tn?2na1?a2???an,求证:T1?T2?T3???Tn?3.

2n解析:Tn?41?42?43???4n?(21?22???2n)?4(1?41?4)?2(1?2n)4n?(4?1)?2(1?2n) 1?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

111 从而T1?T2?T3???Tn?3??1??????2?33711?3 ?n?1??2?12?1?2n例7.已知x1?1,x证明:

4n?n(n?2k?1,k?Z),求证:1 11???????2(n?1?1)(n?N*)4x?x4x?x4xx?n?1(n?2k,k?Z)23452n2n?11?1421x2nx2n?1?4(2n?1)(2n?1)4n?11??144n2?12?n2?22n,

因为 2n?n?n?1,所以

422nx2nx2n?1?n?n?1?2(n?1?n)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???ln3n2343n?3n?5n?6(n?N*). 6

解析:先构造函数有lnx?x?1?lnx?1?1,从而ln2?ln3?ln4???ln3xxn2343n111 ?3n?1?(????n)233cause1?1???231?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???ln32343nn?3n?1?5n5n?6 ?3n?66?ln3?lnn?2n2?n?1 例9.求证:(1)??2,ln2???????(n?2) ?23n2(n?1) 解析:构造函数f(x)?lnx,得到lnnx?n?lnn2?2n,再进行裂项lnn2n2?1?11,求和后可以得到?1?2n(n?1)n答案

函数构造形式: lnx?x?1,lnn?23n?1?n??1(??2) 例10.求证:1?1???1?ln(n?1)?1?1???1 2n解析:提示:ln(n?1)?lnn?1?nn2n?1n????ln?ln???ln2 n?11nn?1y函数构造形式: lnx?x,lnx?1?1 x当然本题的证明还可以运用积分放缩 如图,取函数f(x)?1, xEFDCBnx首先:SABCF1??xn?in,从而,1?i?n1 ?lnx|nn?i?lnn?ln(n?i)?xn?inOAn-i取i?1有,1?lnn?ln(n?1), n所以有1?ln2,1?ln3?ln2,…,1?lnn?ln(n?1),23n1?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,ABDE11,从而有1 ?i???lnx|n??n?i?lnn?ln(n?i)n?in?ixn?ixnn1?lnn?ln(n?1), n?1111 ?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?1)???(1?2!3!n!9811)?e32n.解析:构造函数后即可证明