高考数学 数列压轴题放缩法技巧
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?1n24k?124n?1122的值; (2)求证:??415?. 23k?1kn解析:(1)因为 (2)因为
1?n2211,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?2???2n?12n?1?33?35k?1k14n?1?2n?12n?1?n2?4奇巧积累:(1)
1441? ?1???2???n24n24n2?1?2n?12n?1? (2)
1211 ???12Cn?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( (9)
n?1?n)?1n (8)
1?111?2?????n?n?1(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 ??(n?1)!n!(n?1)!
222n?11?n?22 (10) (11)
1n
?2(2n?1?2n?1)?2n?1?2n?1? (11) (12) (13) (14) (15)
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?1
1?n?1?n?1?1??????n?1?2n?n?111 ?n?1n?1
2n?1?2?2n?(3?1)?2n?3?3(2n?1)?2n?2n?1?k?211 ??k!?(k?1)!?(k?2)!(k?1)!(k?2)!2n12n?n?32?13 (15)
i?ji?1?21?n?n?1(n?2) n(n?1)i2?1?j2?1i2?j2?i?j(i?j)(i2?1?j?1)2?j?12?1
例2.(1)求证:1?11171?2?????(n?2) 2262(2n?1)35(2n?1)1 / 26
高考数学 数列压轴题放缩法技巧
(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(解析:(1)因为 (2)1?4n?1?1)?1?12?13?2(2n?1?1)
n111?11??????(2n?1)2(2n?1)(2n?1)2?2n?12n?1?,所以
?(2i?1)i?112111111 ?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
而由均值不等式知道这是显然成立的，所以
2(n?1?1)?1?12?13???1n再证
1n?2(2n?1?2n?1)?222n?1?2n?1?n?211?n?221?12?13???1n?2(2n?1?1)
例3.求证:
6n1115?1?????2?
(n?1)(2n?1)49n31?n21n2?141??1?2???24n?1?2n?12n?1?4 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,
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)49n,
,所以综上有
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?ak≥a1?b.证明:a?b. k?1a1lnb11)，整数?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
2 / 26
高考数学 数列压轴题放缩法技巧
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???nm,求证:
解析:首先可以证明:(1?x)n?1?nx
nnm?1?(m?1)Sn?(n?1)m?1?1.
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
nk?1nm?1?(m?1)Sn?(n?1)m?1?1只要证:
n
n?[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],即等价于
k?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???Tn?3.
na1?a2???an2解析:Tn?41?42?43???4n?(21?22???2n)?4(1?41?4n)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
?32n3?11????n?n?1?nn2(2?2?1)(2?1)2?2?12?1?
11?3 ?n?1??2?12?1?2n111 从而T1?T2?T3???Tn?3??1??????2?337例7.已知x1?1,x14n?n(n?2k?1,k?Z),求证: ???n?1(n?2k,k?Z)14x2?x3?14x4?x5????x2nx2n?11?2(n?1?1)(n?N*)14
1?22n证明:
41x2nx2n?14(2n?1)(2n?1)?14n2?1?2?144n2?2,因为
2?n 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???ln3n2343n?3n?5n?6(n?N*). 63 / 26
高考数学 数列压轴题放缩法技巧
