高考数列放缩法技巧全总结VIP免费

高考数学备考之放缩技巧证数列型不等式，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：通过多角度观察通项的结构，剖析特征，抓住其规律放缩；其放缩技巧主要有以下几种：一、裂项放缩例1.(1)求n4kk122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14n1111251,所以11211222n12n13335k1k14n21n22n12n1n24注意术语，如求和，奇数技巧积累:(1)1422n4n2111(2)112124n12n12n1Cn1Cn(n1)n(n1)n(n1)n(n1)42(3)Tr1rCn1n!11111r(r2)rr!(nr)!nr!r(r1)r1rn(4)(11)n111n2111532n(n1)21n2nn2(5)111(6)nnnn2(21)21211(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1n(n2)n2nnnnnn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)1n31n1n1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)k211(15)k!(k1)!(k2)!(k1)!(k2)!1nn1(n2)n(n1)ij1(15)i21j21i2j2ij(ij)(i21j1)2i12j12171例2.(1)求证:111(n2)22262(2n1)35(2n1)(2)求证:1111112416364n24n(3)求证:113135135(2n1)2242462462n2n11(4)求证：2(n11)11112(2n11)23n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2n2n1n12n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),所以容易经过裂项得到1n2(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12131n2(2n11)例3.求证:6n111512(n1)(2n1)49n31n21411224n12n12n14解析:一方面:因为1n2,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a1lnb明:ak1b.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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]...