2010高考数学备考之放缩技巧
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?1n224k?12211,所以n212n 解析:(1)因为2????1???24n?1(2n?1)(2n?1)2n?12n?12n?12n?1k?14k?11?n214的值; (2)求证:?15?. 23k?1kn (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?421111? (2)1奇巧积累:(1)1?4?4?2? ??????12222n4n4n?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)2
1?n?2?n n?2 (5)
111?n?nnn2(2?1)2?12 (6)
?2(n?n?1)
(7)2(
1?111?2 ?????n?n?1n(2n?1)?2(2n?3)?2n?2n?12n?3?2111?111?11?? (9)????,????k(n?1?k)?n?1?kk?n?1n(n?1?k)k?1?nn?1?k?n?1?n)?1 (8)
(10) (11) (12) (13) (14) (15)
n11 ??(n?1)!n!(n?1)! (11)
1n?2(2n?1?2n?1)?222n?1?2n?1?n?211?n?222n2n2n2n?111 ????n?1?n(n?2)n2nnnnnn?1(2?1)(2?1)(2?1)(2?1)(2?2)(2?1)(2?1)2?12?11?1?
??1111 ?????n(n?1)(n?1)?n(n?1)?n3n?n2?n(n?1)?n?1?n?11?n?1?n?111 ?1???????n?1?2nn?1n?1?n?12n12n 2?2?2?(3?1)?2?3?3(2?1)?2?2?1??n?32?131k?211 (15) ?n?n?1(n?2) ??k!?(k?1)!?(k?2)!(k?1)!(k?2)!n(n?1)n?1nnnnn
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
4163624n4n(3)求证:1?1?3?1?3?5???1?3?5???(2n?1)?2n?1?1
22?42?4?62?4?6???2nn?1?1)?1?12?13???1n?2(2n?1?1)
(4) 求证：2(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1n?2(n?1?n)?2n?1?n,所以容易经过裂项得到
2(n?1?1)?1?12?13???1n
而由均值不等式知道这是显然成立的，所
?n?211?n?22再证
1n?2(2n?1?2n?1)?222n?1?2n?1以1?12?13???1n?2(2n?1?1)
例3.求证:
6n1115?1?????2?
(n?1)(2n?1)49n31?n21n2??1??1?2???24n?1?2n?12n?1?4 解析:一方面:因为,所以
14n111?25 ?11?1?2????????1???22n?12n?1?33?35k?1k 另一方面:1?1?1??49?11111n ?1??????1??2n2?33?4n(n?1)n?1n?1 当n?3时,
当n?2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)49n6n1115?1?????2?
(n?1)(2n?1)49n3,所以综上有
例4.(2008年全国一卷) 设函数f(x)?x?xlnx.数列?an?满足0?a数k≥a1?b.证明:ak?1?b.
a1lnb1整1)，?1.an?1?f(an).设b?(a1， 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),
kkm?1m?1于是ak?1?a1?k|a1lnb|?a1?(b?a1)?b 例5.已知n,m?N?,x??1,Sm?1m?2m?3m?? 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??
nk?1?nm,求证: nm?1?(m?1)Sn?(n?1)m?1?1.
?1m?1?0??[km?1?(k?1)m?1]所以要证
nk?1nm?1?(m?1)Sn?(n?1)m?1?1只要证:
nk?1
nk?1?[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]故只要证?[km?1?(k?1)m?1]?(m?1)?km??[(k?1)m?1?km?1],即等价于
nnnk?1k?1k?1
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
kkkk而正是成立的,所以原命题成立.
2n例6.已知an?4n?2n,T?,求证:T1?T2?T3??na1?a2???an?Tn?3.
2解析:Tn?41?42?43??所以
2n?4n?(21?22???2n)?4(1?4n)2(1?2n)4n??(4?1)?2(1?2n) 1?41?232n2n3?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?133333n323?11?????n?n?1?nn2(2?2?1)(2?1)2?2?12?1?
从而T1?T2?T3??例7.已知x1?1,x14?Tn?3?11111?3 ?n?1?1??????n??2?3372?12?1?2n?n(n?2k?1,k?Z),求证: ???n?1(n?2k,k?Z)???14x2?x3?14x4?x5x2nx2n?1?2(n?1?1)(n?N*)
证明:
41x2nx2n?1?14(2n?1)(2n?1)?1144n?1?22?144n2?212?n?22n,因为
2n?n?n?1,所以
4x2nx2n?12n?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??234xln3n5n?6?n?3n?(n?N*). 36n 解析:先构造函数有lnx?x?1?lnx?1?1,从而ln2?ln3?ln4???ln3x2343n111 ?3n?1?(????n)233因为1?1???231?11??111111?11? ?1????????????????n?n???n?n32?13??23??456789??2
??3n?15?33??99?3n?1?5n ??????????????2?3n?13n???66?69??1827???n5n5n?6 nn所以ln2?ln3?ln4???ln3?3?1??3?234663n???2ln2ln3lnn2n?n?19.求证:(1)??2,????????(n?2) 2(n?1)23n? 例 解析:构造函数f(x)?lnx,得到lnnxn?lnn2?2n,再进行裂项lnn2n2?1?11,求和后可以得到答?1?2n(n?1)n案 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?11nn?1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:SABCF1??xn?inn,从而,1?i?nnn?i?x?lnx|,…,1nn?i?lnn?ln(n?i) EFDC取i?1有,1?lnn?ln(n?1), 所以有1?ln22,111?????ln(n?1) 23n?11?ln3?ln231?lnn?ln(n?1)n,O1n?1BAn-inlnn?ln(n?1)?,相加x后可以得到: 另一方面S取i?1有,ABDE1??n?ixn,从而有11 ?i???lnx|nn?i?lnn?ln(n?i)n?in?ixn1?lnn?ln(n?1), n?12n所以有ln(n?1)?1?1???1,所以综上有1?1???23111 ?ln(n?1)?1????n?12n811)?e32n例11.求证:(1?1)(1?1)???(1?1)?e和(1?1)(1?1)???(1?2!3!n!9. 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:ln[n(n?1)?1]?2?3,叠加之后就可以得到答案
n(n?1)?1 函数构造形式:ln(x?1)?2?331?ln(1?x)3(x?0)??(x?0)x?1xx?1(加强命题) 例13.证明:ln2?ln3?ln4???lnn45n?1?n(n?1)(n?N*,n?1)4 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?x?1x?1,令f'(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,所以ln2?ln3?ln4???lnn?n(n?1)(n?N*,n?1)
n?1?2345n?14 例14. 已知a 解析:
an?1?(1?1?1,an?1?(1?11证明a?e2. n)a?.nn2?n2n1111)an?n?(1??n)ann(n?1)n(n?1)22,
n?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： 1111lna?ln(1??)?lna?an?1?(1?n2?n?2n)an?n?1n2?n2nn?lnan?11?nn?n22n?1i?1。于是lnan?1?lnan?11?nn?n22，
?i?1n?1(lnai?1?lnai)??11?()n?1111112(2?)?lnan?lna1?1???2??n?2.1nn2i?i2i1?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)?n?1ln(an?1?1)?ln(an?1)?ln(1?11)?.n(n?1)n(n?1)
??[ln(ai?1?1)?ln(ai?1)]??i?2i?2n?1， 11?ln(an?1)?ln(a2?1)?1??1i(i?1)n是在(0,??)上处处可导的函数,若
即ln(an?1)?1?ln3?an?3e?1?e2.
f(x) 例15.(2008年厦门市质检) 已知函数
x?f'(x)?f(x)在x?0上恒成立. (I)求证：函数上是增函数； f(x)g(x)?x在(0,??) (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：1ln2?1ln3?1ln4???1ln(n?1)?n2222223242(n?1)22(n?1)(n?2)g(x)?(n?N*).
解析:(I)g'(x)?
xf'(x)x?f(x)?0x2,所以函数
f(x)在(0,??)x上是增函数
(II)因为g(x)?f(x)在(0,??)上是增函数,所以
f(x1)f(x1?x2)x1 ??f(x1)??f(x1?x2)x1x1?x2x1?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)? (3) f(x)?f(x?x???x)?f(x)?x112n1f(x1?x2)
?f(x1?x2???xn)
x1x1?x2???xn1x1?x2???xnf(xn)f(x1?x2???xn)xn ??f(xn)??f(x1?x2???xn)xnx1?x2???xnx1?x2???xnf(x2)f(x1?x2???xn)x2…… ??f(x2)??f(x1?x2???xn)x2x1?x2???xnx1?x2???xn
相加后可以得到:
f(x1)?f(x2)???f(xn)?f(x1?x2???xn)
?xnlnxn?(x1?x2???xn)ln(x1?x2???xn)
所以x1lnx1?x2lnx2?x3lnx3??有
令
xn?1(1?n)2,
?11112222????22ln2?32ln3?42ln4???(n?1)2ln(n?1)??????1111??111???22?32?42???(n?1)2???ln??22?32???(n?1)2??????
?111??111????2?1?3?2???(n?1)n???22?32???(n?1)2???ln?????
1??11?n? ?????????2(n?1)(n?2)?n?1??2n?2?1111nln22?2ln32?2ln42???ln(n?1)2?222(n?1)(n?2)234(n?1)2 所以
(n?N*).
(方法二)ln(n?1)(n?1)2?ln(n?1)2ln41? ?1??ln4???(n?1)(n?2)(n?1)(n?2)?n?1n?2? 所以
11111?nln4?1ln22?2ln32?2ln42???ln(n?1)2?ln4????22234(n?1)?2n?2?2(n?2)
又ln4?1?1,所以1 111nln22?2ln32?2ln42???ln(n?1)2?(n?N*).22n?12(n?1)(n?2)234(n?1) 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) ?f(x)?xlnx,
?g(x)?xlnx?(k?x)ln(k?x),?0?x?k.?g?(x)?lnx?1?ln(k?x)?1?lnx,k?xx2x?kk令g?(x)?0,则有?1??0??x?k.k?xk?x2kg(x)在[,k2 ∴函数
）上单调递增，在(0,k]上单调递减.
2kg()2∴g(x)的最小值为
222，即总有g(x)?g(k).
22而g(k)?f(k)?f(k?k)?klnk?k(lnk?ln2)?f(k)?kln2,
?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).
三、分式放缩
姐妹不等式:b?b?m(b?a?0,m?0)和b?b?m(a?b?0,m?0)
aa?maa?m 记忆口诀”小者小,大者大”
解释:看b,若b小,则不等号是小于号,反之. 例19. 姐妹不等式:(1?1)(1?1)(1?1)?(1?1)?2n?1和
3512n?112n?1aa?m2n?11111(1?)(1?)(1?)?(1?)?2462n也可以表示成为
和1?3?5???(2n?1)?2?4?6??2n?2n?12?4?6???2n1?3?5???(2n?1) 2?4?6?2n?3?5?7?1352n?1246?(2?4?6?解析: 利用假分数的一个性质b?b?m(b?a?0,m?0)可得
2n?11352n?1 ?????(2n?1)2n2462n1351112n2)?2n?1. )?2n?1即(1?1)(1?)(1?)?(1?352n?12n?1 例20.证明:(1?1)(1?1)(1?1)?(1?471)?33n?1. 3n?2
147258????147解析: 运用两次次分式放缩: 2583n?13693n (加1) ??????.?????3n?22583n?13n?147103n?1 ??.?????3n?23693n2 (加2)
相乘,可以得到:
3n?1?47103n?11473n?2?258 ???????(3n?1)????????.?????3n?2?2583n?12583n?1?147
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