數(shù)列中的存在性問題 經(jīng)典(教師)

.naSn25nbb164bbn例nn=nn1nalogbc，nMM和cncnalogbc，nMn，ncSn25n，n=∵aS11=naSS3562n2nn2nnn當(dāng)n=nn1==，n當(dāng)a62n∴n=，bn1164bbb64n1n∴=，n1bn1)n1b=2，96n∴則n=alogb6n2296n6n2(96n)log26(1log2)29log2n，ancn=c==aaalogbnMn，，nc6(1log2)c∴a29log2M∴=aalogbc，nMnnc..例a}S，aa34，S9n且．nn5133a}nnabnb}atb，b，bn1nn2m(mmN)t和maa34，513a，a}8d17，分2naa，11ad，d2.即故1a2n，Sn2分nn2n1b，b，bbbb2n1t.要使b1成等差數(shù)列，必須，即mn（1）知12m2132m123t1t2m1t4m3t111分t當(dāng)t2m7t3m5t5m4.b，b，b115分2mabbn(mN*)naSn2amn例.nnnnb,b,bm128b,,(,5)bbbtNb*tmtn14tmSn2aSS2n1n2nnn1分naS1a2n1n111n（nN*）…4分..2n113bb,b,bb22bb2n1m，則11m3mmn28，1831()20m9m9分3m1mm得mbbbb1,b,b(tN,t5)*mt4t4171t12t77m1mt1mm5分m51,2,3,4,6,9,12,18,36t43,25,19,16,13,11,10,9,8mm9個分例aSnaS20nn2n1n1baabnTn.nn1nnabTn；nnn,n(1mn)1,,,nmnTTT(aa)(2naSna(22121na2n2n1n，na0a2n1nn分1111((2nn22n12n11b)aann1n由111111nTL)23352n12n12n1n分n1mnTT,T,T32m12n1，nmn2n1，1m1n)(m2n()2T,T,T21m32n14m4m16n32若1mn分..m2n32m4m124m4m16n3nm22，2m4m102，分16m1622mNm1m2n12．m2n12，TT,T,Tn1mn分n13m216n364m4m162m4m10n22分6m16122例a}apq(p0,q0,pq,R,0,nN*)nnn，n.apa}n1na}nA{(n,b)|b3k,nN*}kN，nnknnB{(n,c)|c5,nN*}nnnapapn1qn1ppqqqp)(n)(nnapnnqn=n1n，aqn1n2∵∴0,q0,pq∴an1napa}數(shù)列n1n為等比數(shù)列a}a,a,a(n1,nN)nnn1n2，..aa∵a2n1nn2q12pqpn2qn2pq(pq(pn1)()()nnnnn)2，Qp0,qpq,0pq(pq)0a1aann2nn2，a}∴數(shù)列n中不存在連續(xù)三項構(gòu)成等比數(shù)列；BIC；k13k315nnnn當(dāng)k33k3323BIC；n5nnnnnk53k5BIC當(dāng)時，nnn，此時；當(dāng)k2325nnnn1n132()()1nn32555由nnn得，3232f(x)()()f(x)()()xxxx5555設(shè)∴是Rf(x)132()()1nnBIC155325n時n2n2；nnn當(dāng)k4345nnn34()()1nnBICn2時55345；nnnk1k3k5BIC當(dāng)k2BIC{(1,5)}當(dāng)k4，或；，BIC{(2,25)}。..例1a+,n為偶數(shù),4a=21n2a-a+,為奇數(shù),2a}N,aR,a*1nn（，2b}bann。221na,a,a123；b}nb}an11a2aaaa11122，114aaa421，1a2aaa232。分ba2aa1⑵n2，n221n221111132baaaa)aaa1a)a1aa2424n1221n21n2n2nn21n21bb1baa∴n1n13，b}a1分nnban1，分ai,aj,ak、、ijkjik(ai)(aj)(ak)a(ik2j)jik，22分..若ik2j0j2ik0==ijk分jik2jika若ik2j0ik2j，jik∵、、a∴分例nnn故＝2)．nb2bbprq2)，∴∴2例a(ij)ijj,ijN*i表示第行第個數(shù)（由部分自然數(shù)構(gòu)成如圖的數(shù)表，用aai(nN*)ni1iibn。b6；bn1；n..b}bpqrN,,*bbn，，（pqpq，，rb946分ban1a...a(n1)1(n1)2(n1)(n1)n1(aa)...(aa)n1n(n1)nnn1n22(aa...a)2n1n2nnb2=n分bb2b22(b2)n1nn1n分b2}b231n分b232nnb32n2.分11則n,b,b(p,q,rN*)b}n中存在不同的三項bpqrb}bbbpqrn分rqp2(322)(322)(322)q1p1r1即22221qrpr分p,q,rN*pqrqrprb,b,b(p,q,rN*)b}故數(shù)列n中不存在不同的三項恰好成等差數(shù)pqr分..例3+2naa=Nn–1nnana+pa–2nb=bnnn,n,pN,mnpm*a，，，，是aaaaaanmnpmnp4解a=2+naa=分n–11nn42++pn–14+–+4++–)nnb===b44nnn–1則b–bb=nN++(2––++(2–+n22n+–=N–p–3–3)=0即––p=p=2n2n7分p=2b=3bp=–2b=bnn=n±nn為當(dāng)p2時b}等比數(shù)n分4443p142，a2，a2aa，，，aaamm1n1mnpnpm442aaa)=22，apan1m13p1nnmp分3(2331)1323npnpmpmnmm*,n,pN,mnp，，mpn1pmnm1，p3333pn33nm133nm，pmpn1pm3(23331)3(31)0，npnpnnpn13323130nmnmnma，，，，分aaaaaanmnpmnp例a3an1,n1,2,Lana2a115，．nn11an..111SLS100naa1ann．記n2,s,n,s,na1,a1,a1msn1213a1111aaa3分n1nn1n110110(nN*)aa，分1n11a分n121111()2()1n1na33a35分nn11n11133n2n11111111nSLnL)aa1a3233nn分2n1n1100S100maxn99分3n若nmn2s,(a1)(a1)(a1)2mns，分nnm3sa(1)((2n2n2m23s2n∵分3323mns分332323mnmnmns∵分又,,s分例a}SaSnN)2n*nn1nn..a}nan2danan與a1n2與n11入1da23與2ag(n)dAnnn.2Ag(n)dn得nnN*d，d，dL，dLd，d，dp123nmk，，paq2a211aaqn1na2S2(nN*)aq22(aaq)21分n1n111a21q3a23nn1，∴分232343n1(2323)(nnn1nn1dnAn4