高考数学-压轴题-放缩法技巧全总结最强大VIP免费

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124n211n2n的值;(2)求证:15.2k1k3解析:(1)因为211,所以n212n12(2n1)(2n1)2n12n12n12n1k14k14(2)因为n1111251,所以112112222n12n13335k1k14n12n12n1n241奇巧积累:(1)144112222n4n4n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11n213215n(n1)21n2nn22n12n3211n1(2n1)2(2n3)2n(5)111nnnn2(21)212(6)211(7)2(n1n)12(nn1)(8)nn(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)!(10)(11)1n2(2n12n1)222n12n1n211n22(11)(12)(13)(14)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11n1n11n12nn12n1nnn11n1n1nn1nn1(n2)n(n1)2n12n22(31)233(21)221n3213k211k!(k1)!(k2)!(k1)!(k2)!(15)22(15)i1j1i2j2(ij)(i12ijj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1112416364n24n(3)求证:113135135(2n1)2n112242462462nn(4)求证：2(n11)11112(2n11)23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2n12n1,再结合1n2n2n进行裂项,最后就可以得到答案(4)首先再证1n1n2(n1n)2n1n22,所以容易经过裂项得到2(n11)111123n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n211n22所以11112(2n11)23n例3.求证:6n111512(n1)(2n1)49n31n2112214n12n12n12n414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1111249n233411n1n(n1)n1n1当n3时,当n2时,所以综上有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例4.(2008年全国一卷)设函数f(x)xxlnx.数列an满足0a明:ak1b.11)，整数k≥a1b1.an1f(an).设b(a1，.证a1lnb解析:由数学归纳法可以证明an是递增数列,故若存在正整数mk,使amb,则ak1akb,若amb(mk),则由0a1amb1知amlnama1lnama1lnb0,aaalnaakalna,mmk1kkk1m1因为am1kmlnam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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n[km1(k...