loading高考数学备考之放缩技巧专题
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
?4kk?1n22的值; (2)求证:
?1??kk?1n12?5. 3解析:(1)因为 (2)因为124n2?11211,所以n212n ???1???2(2n?1)(2n?1)2n?12n?12n?12n?1k?14k?1n1111?25 1?,所以?1?1?2??1????????1??2???2???2n?12n?1?33?35k?1k14n2?1n2?2n?12n?1?n2?44奇巧积累:(1)1?4?22n4n2111? (2)1?1 ????2???1224n?1?2n?12n?1?Cn?1Cn(n?1)n(n?1)n(n?1)n(n?1)4 (3)Tr?1r?Cn?1n!11111??????(r?2) rrr!(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?n?n?n?1?n(n?2)n2nnn?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?j2?1)?i?ji2?1?j2?1?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???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1n?2(n?1?n)?2n?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?1?1???1?2(2n?1?1)
23n例3.求证:
6n1115?1?????2?
(n?1)(2n?1)49n31??1??2?2???21n4n?1?2n?12n?1?n2?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:ak?1?b.
a1lnb 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]k?1k?1k?1nnn 故只要证
?[kk?1nm?1?(k?1)m?1]?(m?1)?k??[(k?1)m?1?km?1],即等价于
mk?1k?1nnkm?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??4n44424?3?2n?1?222?(2n)2?3?2n?1(4?1)?2(1?2n)??2?2n?1??2n?1333332n
?32n3?11? ?????2(2?2n?1)(2n?1)2?2n?12n?1?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?x314?14x4?x5????114x2nx2n?1??2(n?1?1)(n?N*)14
证明:
x2nx2n?14(2n?1)(2n?1)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)
1x2?x3?14x4?x5???14x2nx2n?1?2(n?1?1)(n?N*)
二、函数放缩
n 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)
nnxx2343233
因为112?3???1?11??111111?11?1????????????????n?n???nn32?13?23??456789??2?
???3n?15?33??99?3n?1?5n
??????????????2?3n?1?3n???66?69??1827???n所以ln2?ln3?ln4???ln3?3n?1?5n?3n?5n?6
234663n???2 例9.求证:(1)??2,ln2?ln3???lnn?2n?n?1(n?2)
???23n2(n?1) 解析:构造函数答案 ?lnn2lnx,得到lnn?f(x)?n?n2x2,再进行裂项lnn?1?1?1?1,求和后可以得到n(n?1)n2n2 函数构造形式: lnx?x?1,lnn??n??1(??2) 例10.求证:1?1???123n?1?ln(n?1)?1?11??? 2nn?1解析:提示:ln(n?1)?lnn?1?n???2?lnn?1?lnn???ln2 nn?11n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, xy首先:SABCF11,从而,1 ?i???lnx|n??n?i?lnn?ln(n?i)nxxn?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?in?ixn?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)?1(加强命题) 31?ln(1?x)3(x?0)??(x?0)x?1xx?1ln[n(n?1)?1]?2? 函数构造形式:ln(x?1)?2?
例13.证明:ln2ln3ln43?4?5??? lnnn(n?1)?(n?N*,n?1)n?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 解析:
a1?1,an?1?(1?11证明a?e2. n)an?n.n?n22an?1?(1?, 1111)an?n?(1??n)ann(n?1)n(n?1)22lnan?1?ln(1?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?11?)an?n?n2n2lnan?1?ln(1? 11?n)?lnan?n?n22?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?i2 11?()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：
11)an??an?1?1?(1?1)(an?1)?n(n?1)n(n?1)n(n?1)11ln(an?1?1)?ln(an?1)?ln(1?)?.n(n?1)n(n?1)an?1?(1?
??[ln(ai?1?1)?ln(ai?1)]??i?2i?2n?1n?1， 11?ln(an?1)?ln(a2?1)?1??1i(i?1)n即ln(an?1)?1?ln3?an?3e?1?e2.
例15.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,??)x上是增函数；
(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) (II)因为
g'(x)?,所以函数上是增函数 f'(x)x?f(x)f(x)?0g(x)?在(0,??)2xxg(x)?上是增函数,所以 f(x)在(0,??)xx1 ?f(x1?x2)x1?x2 f(x1)?f(x1?x2)?f(x)?1x1x1?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)