高考数学_压轴题_放缩法技巧全总结

2011高考数学备考之放缩技证明列型等式，因其思维跨度大构造性强，需要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后继学习能力，因而高考压轴题及各级各类竞赛试题命题的极好素材类问题的求解策略是通过多角度观察所给列通项的结构，深入剖析其特，抓住其规律进行当地放缩其放缩技巧要以下几种一裂放缩例1.(1)求∑=−nkk12142的值;(2)求证:35112∑=nkk.解析:(1)因为121121)12)(12(21422+−−=+−=−nnnnn,122121114212+=+−=−∑=nnnknk(2)因为+−−=−=−12112121444111222nnnnn,35321121121513121112=++−−++−+∑=nnknkL奇积累:(1)+−−=−=1211212144441222nnnnn(2))1(1)1(1)1()1(21211+−−=−+=+nnnnnnnCCnn(3))2(111)1(1!11)!(!!11≥−−=−⋅−=⋅=+rrrrrrnrnrnnCTrrrnr(4)25)1(123112111)11(−++×+×+++nnnnL(5)nnnn21121)12(21−−=−(6)nnn−++221(7))1(21)1(2−−−+nnnnn(8)nnnnnnn2)32(12)12(1213211221⋅+−⋅+=⋅+−+−(9)++−+=++++−+=−+knnkknnnkknknk11111)1(1,11111)1(1(10)!)1(1!1!)1(+−=+nnnn(11)21212121222)1212(21−++=−++=−−+nnnnnnn(11))2(121121)12)(12(2)22)(12(2)12)(12(2)12(21112≥−−−=−−=−−−−=−−−−nnnnnnnnnnnnnn(12)111)1(1)1(1)1)(1(11123−−+⋅+−−=+−⋅=nnnnnnnnnnnn11112111111+−−−++⋅+−−=nnnnnnn(13)3212132122)12(332)13(2221nnnnnnnnn−⇒−⇒−⇒⋅−=⋅=+(14)!)2(1!)1(1)!2()!1(!2+−+=+++++kkkkkk(15))2(1)1(1≥−−+nnnnn(15)111)11)((1122222222++++=+++−−=−+−+jijijijijijiji例2.(1)求:)2()12(2167)12(151311222≥−−−++++nnnL(2)求:nn412141361161412−++++L(3)求:1122642)12(531642531423121−+⋅⋅⋅⋅−⋅⋅⋅⋅++⋅⋅⋅⋅+⋅⋅+nnnLLL(4)求)112(2131211)11(2−+++++−+nnnL解析:(1)因+−−=+−−12112121)12)(12(1)12(12nnnnn,所)12131(211)12131(211)12(112−−++−+−∑=nnini(2))111(41)1211(414136116141222nnn−++++=++++LL(3)先运用式放缩法明出1212642)12(531+⋅⋅⋅⋅−⋅⋅⋅⋅nnnLL,再结合nnn−++221进裂,最后就可得到答案(4)首先nnnnn++=−+12)1(21,所容易过裂得到nn131211)11(2++++−+L再21212121222)1212(21−++=−++=−−+nnnnnnn而由均值等式知道是显然成立的所)112(2131211−+++++nnL例3.求:35191411)12)(1(62++++≤++nnnnL解析:一方面:因+−−=−=−12112121444111222nnnnn,所35321121121513121112=++−−++−+∑=nnknkL另一方面:1111)1(143132111914112+=+−=+++×+×+++++nnnnnnLL3≥n时,)12)(1(61+++nnnnn,1=n时,2191411)12)(1(6nnnn++++=++L,2=n时,2191411)12)(1(6nnnn++++++L,所综有35191411)12)(1(62++++≤++nnnnL例4.(2008年全国一卷)设函数()lnfxxxx=−.数列{}na满足101a.1()nnafa+=.设1(1)ba∈整数11lnabkab−.明:1kab+.解析:由数学纳法可明{}na是递增数列,故若存在整数km≤,使bam≥,则baakk≥+1,若)(kmbam≤,则由101≤baam知0lnlnln11≤baaaaammm,∑=+−=−=kmmmkkkkaaaaaaa111lnln,因)ln(ln11bakaakmmm∑=,于是bababakaak=−+≥++)(|ln|11111例5.已知mmmmmnSxNmn++++=−∈+L321,1,,,求:1)1()1(11−++++mnmnSmn.解析:首先可明:nxxn+≥+1)1(∑=++++++++−−=−++−−−+−−=nkmmmmmmmmkknnnnn111111111])1([01)2()1()1(L所要1)1()1(11−++++mnmnSmn只要:∑∑∑=+++++++++==++−+=−++−−+−+=−++−−nkmmmmmmmmmnkmnkmmkknnnnnkmkk111111111111111])1[(2)1()1(1)1()1(])1([L故只要∑∑∑=++==++−++−−nkmmnkmnkmmkkkmkk1111111])1[()1(])1([,即等于mmmmmkkkmkk−++−−+++111)1()1()1(,即等于11)11(11,)11(11++−+−+++mmkkmkkm而是成立的,所原命题成立.例6.已知nnna24−=,nnnaaaT+++=L212,求:23321++++nTTTTL.解析:)21(2)14(3421)21(241)41(4)222(444421321nnnnnnnT−+−=−−−−−=+++−++++=LL所123)2(22232234232323422234342)21(2)14(3422111111+⋅−⋅⋅=+⋅−⋅=−+=−+−=−+−=++++++nnnnnnnnnnnnnnnnT−−−=−−⋅⋅=[phone])(122(2231nnnnn从而231211217131311231321−−−++−+−=+++++nnnTTTTLL例7.已知11=x,∈=−∈−==),2(1),12(ZkknnZkknnxn,求:*))(11(21114122454432Nnnxxxxxxnn∈−+++⋅+⋅+L明:nnnnnnxxnn222141141)12)(12(11424244122=⋅=−=+−=+,因12++nnn,所)1(2122214122nnnnnxxnn−+=+++所*))(11(21114122454432Nnnxxxxxxnn∈−+++⋅+⋅+L二函数放缩例8.求)(665333ln44ln33ln22ln*Nnnnnn∈+−++++L.解析:先构造函数有xxxxx11ln1ln−≤⇒−≤,从而)313121(1333ln44ln33ln22lnnnnn+++−−++++LLcause+++++++++++++=+++nnnn31121219181716151413121313121LLL6533323279189936365111nnnnn=+⋅++++++−−−L所6653651333ln44ln33ln22ln+−=−−++++nnnnnnL例9.求:(1))2()1(212ln33ln22ln,22≥+−−+++≥nnnnnnαααααααL解析:构造函数xxxfln)(=,得到22lnlnnnnn≤αα,再进裂)1(1111ln222+−−≤nnnnn,求和后可得到答案函数构造形式:1ln−≤xx,)2(1ln≥−≤αααnn例10.求:nnn1211)1ln(113121++++++++LL解析:提示:2ln1ln1ln1211ln)1ln(++−++=⋅⋅−⋅+=+LLnnnnnnnnn函数构造形式:xxxx11ln,ln−然本题的明可运用放缩如图,取函数xxf1)(=,首先:∫−ninABCFxS1,从而,)ln(ln|ln11innxxinninnin−−==⋅−−∫取1=i有,)1ln(ln1−−nnn,所有2ln21,2ln3ln31−,…,)1ln(ln1−−nnn,nnnln)1ln(11−++,相加后可得到:)1ln(113121+++++nnL另一方面∫−ninABDExS1,从而有)ln(ln|ln11innxxiinninnin−−==⋅−−−∫取1=i有,)1ln(ln11−−−nnn,所有nn1211)1ln(++++L,所综有nnn1211)1ln(113121++++++++LL例11.求:en+⋅⋅++)!11()!311)(!211(L和en+⋅⋅++)311()8111)(911(2L.解析:构造函数后即可明例12.求:32)]1(1[)321()211(−++⋅⋅×+⋅×+nennL解析:1)1(32]1)1(ln[++−++nnnn,叠加之后就可得到答案函数构造形式:)0(13)1ln(1)0(132)1ln(+++⇔+−+xxxxxxx(加强命题)例13.明:)1*,(4)1(1ln54ln43ln32ln∈−+++++nNnnnnnL解析:构造函数)1(1)1()1ln()(+−−−=xxxxf,求,可得到:12111)('−−=−−=xxxxf,0)('xf有21x,0)('xf有2x,所0)2()(=≤fxf,所2)1ln(−≤−xx,12+=nx有,1ln22−≤nn所211ln−≤+nnn,所)1*,(4)1(1ln54ln43ln32ln∈−+++++nNnnnnnLFEDCBAn-inyxO例14.已知112111,(1).2nnnaaann+==+++明2nae.解析:nnnnnannanna)21)1(11(21))1(11(1++++++=+,然后两边取自然对数,可得到nnnannaln)21)1(11ln(ln1+++++然后运用xx+)1ln(和裂可得到答案)放缩思路⇒+++≤+nnnanna)2111(21⇒++++≤+nnnannaln)2111ln(ln21nnnna211ln2+++≤于是nnnnnaa211lnln21++≤−+.22112211)21(111lnln)211()ln(ln11211111−−=−−+−≤−⇒++≤−−−=+−=∑∑nnniniiininnaaiiaa即.2lnln21eaaann⇒−注题目所给条ln(1)xx+0x一有用结论可起到提醒思路探索放缩方向的作用然本题可用结论)2)(1(2≥−nnnn来放缩⇒−+−+≤+)1(1))1(11(1nnannann⇒+−+≤++)1)()1(11(11nnanna.)1(1))1(11ln()1ln()1ln(1−−+≤+−++nnnnaann111)1ln()1ln()1(1)]1ln()1ln([212112−+−+⇒−+−+⇒∑∑−=+−=naaiiaanniiini即.133ln1)