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?1n24k2?122的值; (2)求证:?n15. ?23k?1k解析:(1)因为
4n?1?1211,所以n212n ???1???2(2n?1)(2n?1)2n?12n?12n?12n?1k?14k?14 (2)因为11??1???2???14n2?1n2?2n?12n?1?n2?4,所以?1n11?25?11?1?2????????1??2n?12n?1?33?35k?1k2
奇巧积累 :
(1)1n2?441??1?2?2???24n4n?1?2n?12n?1?r?1r?Cn? (2)
1211???2CCn(n?1)n(n?1)n(n?1)n(n?1)1n?1
(3)T1n!11111??r????(r?2) rr!(n?r)!nr!r(r?1)r?1rn
(4)(1?1)n?1?1?n1115????? 2?13?2n(n?1)2
1?n?2?n n?2 (5) (7)
111?n?nnn2(2?1)2?12 (6)
1n?2(n?n?1) (8)
2(n?1?n)?
1?111?2?????n?n?1(2n?1)?2(2n?3)?2n?2n?12n?3?2
n11??(n?1)!n!(n?1)! (9)
111?111?11?? ????,????k(n?1?k)?n?1?kk?n?1n(n?1?k)k?1?nn?1?k? (10) (11)
1n?2(2n?1?2n?1)?222n?1?2n?1?n?211?n?22
(11) (12) (13)
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?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?12n?1nnn11?n?1n?1nn
2n12n?2?2?(3?1)?2?3?3(2?1)?2?2?1??n?32?13 (14)
k?211??k!?(k?1)!?(k?2)!(k?1)!(k?2)! (15)
1?n?n?1(n?2) n(n?1) (15) 例2.(1)求证:1?(2)求证:1?42i2?1?j2?1i2?j2?i?j(i?j)(i2?1?j?1)2?i?ji?1?2j?12?1
11171?2?????(n?2) 2262(2n?1)35(2n?1)11111 ????2??163624n4n2?42?4?6
2n?1?1
(3)求证:1?1?3?1?3?5???1?3?5???(2n?1)?2?4?6???2n???1n(4) 求证：2(n?1?1)?1?12?13?2(2n?1?1)
?(2i?1)i?1n解析:(1)因为 (2)1?4111?11??????2(2n?1)(2n?1)2?2n?12n?1?(2n?1),所以
12111111?1?(?)?1?(?)232n?1232n?1
11111111????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:解析:
n6n1115?1?????2?
(n?1)(2n?1)49n3 一方面: 因为
1?n21??1?2?2???14n?12n?12n?1?2?n?414,所以
?kk?11211?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,
n6n?n?1(n?1)(2n?1),当n?1时,
6n111?1?????2(n?1)(2n?1)49n,
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?a?满足0?a?1.an1n?1?f(an).设b?(a，1)，整
1数k≥a?b.证明:a1a1lnbk?1?b.
解析: 由数学归纳法可以证明?a?是递增数列, 故 若存在正整数m?k, 使anm?b,
则ak?1?ak?b,
,则由0?akm1若am?b(m?k)?am?b?1知amlnam?a1lnam?a1lnb?0a?a?alna?a??alnak?1kkk1mmm?1,
k,
因为?am?1lnam?k(a1lnb),于是ak?1?a1?k|a1lnb|?a1?(b?a1)?b
例5.已知n,m?N?,x??1,Sm?1m?2m?3m???nmn,求证:
nm?1?(m?1)Sn?(n?1)m?1?1.
解析:首先可以证明:(1?x)
nk?1?1?nxnm?1?nm?1?(n?1)m?1?(n?1)m?1?(n?2)m?1???1m?1?0??[km?1?(k?1)m?1]k?1n所以要证
nm?1?(m?1)Sn?(n?1)m?1?1n只要证:
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]k?1
故只要证?[knk?1m?1?(k?1)m?1]?(m?1)?km??[(k?1)m?1?km?1]k?1k?1m?1nn,
即等价于k?(k?1)m?1?(m?1)km?(k?1)m?1?kmm?1,
即等价于1?m?1?(1?1)kk,1?m?11?(1?)m?1kk 而正是成立的,所以原命题成立.
例6.已知an?4n?2nT?n1,
2na1?a2???an23,求证:T?T1n122?T3???Tn?n32.
解析:T所以
n4(1?4n)2(1?2n)4n?4?4?4???4?(2?2???2)???(4?1)?2(1?2n)1?41?232n2n2n3?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?133333
?32n3?11????n?n?1?nn2(2?2?1)(2?1)2?2?12?1?
2019年高考数学压轴题放缩法技巧全总结(完整版)
