高中数列放缩法技巧大全VIP免费

高中数列放缩法技巧大全证明数列型不等式，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nn例1.(1)求2125k14k21的值;(2)求证:k1k3.解析:(1)因为22114n21(2n1)(2n1)2n12n1,所以n22k14k11122n1n2n1(2)因为11n2n2144n21212n112n1,4n所以11111125k1k212352n12n133技巧积累:(1)1441n24n24n21212n12n1(2)1211C12(n1)n(n1)n(n1)n(n1)n1Cn(3)TCrr1n1nrn!11111r!(nr)!nrr!r(r1)r1r(r2)(4)(11n)n111211321n(n1)52(5)12n(2n1)12n112n(6)1n2n2n(7)2(n1n)1n2(nn1)111112n(8)n1n2n12n32(2n1)2(2n3)2(9)111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12n1n211n22(10)(11)12(2n12n1)n(12)2n2n2n2n111(n2)n2nnnnnn1n1n(21)(21)(21)(21)(22)(21)(21)2121(13)1n31nn21111n(n1)(n1)n(n1)n(n1)n1n11n1n11n12nn111n1n1(14)2n12n12n22(31)233(21)221n3213nnnnn(15)k211k!(k1)!(k2)!(k1)!(k2)!1nn1(n2)n(n1)(16)(17)i21j21i2j2ij(ij)(i21例2.(1)求证:1j1)2iji12j12111171(n2)22262(2n1)35(2n1)111111(2)求证:24163624n4n2(3)求证:113135135(2n1)2n112242462462n12131n2(2n11)(4)求证：2(n11)1解析:(1)因为n11111,2(2n1)(2n1)22n12n1(2n1)所以i111111111()1()2232n1232n1(2i1)(2)1111111112(122)(11)4163644n4n2n135(2n1)2462n12n1,再结合(3)先运用分式放缩法证明出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)12132n1n1n,所以容易经过裂项得到2(n11)1再证1n2(2n12n1)222n12n1211n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)49n312n1n214112,所以24n12n12n14解析:一方面:因为3111251112122n12n13335k1kn另一方面:11111111n21149n2334n(n1)n1n16n,当n1(n1)(2n1)当n3时,nn1时,6n11112,(n1)(2n1)49n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b.1)，整数k≥0a11.an1f(an).设b(a1，解析:由数学归纳法可以证明a1b.证明:ak1a1lnban是递增数列,故存在正整数mk,使kam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,m1因为amlnamk(a1lnb),于是m1kak1a1k|a1lnb|a1(ba1)b例5.已知n,mN,x1,Sm123n,求证:nm1mmmm(m1)Sn(n1)m11.n解析:首先可以证明:(1x)1nx4nm1nm1(n1)m1(n1)m1(n2)m11m10[km1(k...