高考數(shù)學(xué)北師大理一輪復(fù)習(xí)-第章-數(shù)列-高考專(zhuān)題突破三-高考中的數(shù)列問(wèn)題-文檔

aaaaabann4b1bb3bb52()D2ab2ab＋abaaa11nnn15124aa82．已知等比數(shù)列的各項(xiàng)都為正數(shù)，且當(dāng)n≥3時(shí)，aa＝，則數(shù)列n4n4－aa2a2a2anS()23n14nn·2n2112121nnCa3aaaa，2n44nn2S·2，1na211nnn2，2①②－1n22，S223nnS2·22·2)·2，2－1nnnnnS2n11aaNaaa＋aa1n1＋11n12()Baaaa，11nnaa1naaaaaa)n121，1nn1＝2112－＝.nna1＋n11＋＋1aa1211111＋－＋－＋112－－2233412＝－1anSaS()nn55aan1Aaa.n1a，1，1aaSa＋551＝，，d21aa.n11111∴＝＝－，n1aan111111＋－＋－－＝－.223aan121anSNS＝a－SNk33nn＋nnk＋．4當(dāng)S＝a－2313，11232aa＝a－a，31nna，n1又a，－aa1nnk1S，＝3k由S，kk又N＋3例1anSNSaS＋＋2nn335aSa544an1設(shè)TS－NTSnn＋nn解a，nSSaSaSa3535544aSaSaSaaa，533445553a1q＝＝5.2a433212aa＝.n1an3213－1a＝×·.22nn1.＋21n－＝得S－n21n－2n當(dāng)nS隨nn32S，S＝n111325故S＝－＝－S－.S1S236nn1當(dāng)nS隨nn34SS，2n11347故S＝－－S－.S2S43nn2715S－≤.N＋6Snn567T.n1aaaaan1125an記SanSnnnn，解ad2nd0或2當(dāng)0a當(dāng)4aa；nn1)·4，a2或annn當(dāng)a2S.nn，Sn當(dāng)a2n]S＝n2.2令nn，22或，Snna2；n當(dāng)a2nn121n例2anSa＝a＝a.＋n1nn1nan{nanS.nn1n12aa＝＝a，n11a當(dāng)Nn＋na112aa1又＝，1∶＝1nnnN(∈＋n21112a2nna1212解{nna11n22得＝，·()nn12a·().nn12121212S2·(3·(·(，)))23nn12121212122·(·(，S1·(23))1nnn121122121212∴(((·(S＝n23))1nn11221＝·(，121212S·(()n－n(1).n21212aS·()().nnnnaNababbb，＋n1n2n135bbb135aa；nn1若c＝cnS.bnnnnn解bbb，135aaa，212325aaaa，6213521.1aqbaq22213aba，1121bbbbaa，1355255a1＝q，52a43，aq2a，1111＝，＝.2q4212a5nn．×nnbnn，n知a2－5n11c＝，＝nnnn－n121123113411S－－－－(((n1n111112233411)－＋－＋－＋－n1n1－N．＝111例3Na2＋n11′aa11nan記b＝aabnT.nn1nn解，n，21＝，212則＝N.2＋11a′，an1n又，1111∴＝－，aa11nn11－n，2a4n4a＝，2n2n當(dāng)144a．＝n2n24b＝aa＝nn122nn11－，212nnTbbbn12n＝aa＋aa＋aa1223n111131135－＋－－nn21211－n2n＝.12例4aaaan236anSanST＝Tmnn3·2－nnn1n數(shù)m解a＋，，1ad1a.27d3n＝，ad13S＝，2nTT，＝nn－T＝n1n21，21323T，>TTTT＝n1123Tm32.2n3例5紅，解設(shè)a＋n則，a，2aa．n1aa，n1aa2n1aa，nn，1a2－44－，由a2nnnaab2N．x＋nnn若aabanS；17nn1a若aabx－n122bnnT.n解babab，77887有aaa877aa87S.＋2n1ax2aa2baaa，x22222221xa－.211a－，－2aa21ab2.nnn1123nT＋＋＋＋，＝2222n232n1123＋＋＋nT＝.122n22n1111nT＋＋－2nT＋22nn22n11n－22n12＝.n2T＝.nnaanSSnR．2n2nnλan1＋λbnT.bSnnn解a，naSS，221a，S1211aa，21aa.n1知，1∴b2，－1n1Snn111－b－2，－1nnn＋nn111n1－121123－－∴T＝＋1＋…＋2＋－＋…＋]＝－＝2－)－[－－1nnn1n211.(2015·SanaaaS2nnnnnnan1設(shè)b＝bnnaa＋nn1n解由aaS，2nnn2aS11＋1212aaaa，n1nn1即aaaaaaaa．22n1n11n1naaan1n又aaaaa211111a2ann由a1n11111－＝＝.b＝n212aa22nnnnn1bnTnnTbbbn12n1111111－＝－＋－5735212nnn＝.anSSaN．n＋nnnnaaaa；n12323a＋annn解在SaNn＋nna＝，aa，1111aaa，a＝，01222aaaa，a12333由SaNn＋nnSan11aa，nn14323－aa－nnn1432a＋，1n312323aa＋＋1．nn1232133a＝2＋a－nn1231a＋，nn213212n－n＝－a＝21.nn33n(2015·anSaaaSSNnn122n1.＋aa；2n求S.naSS，2n1NaSS＋11naaaa，2n1n1即a又aa2na，12aSSaaaa，3121121Naa.＋2na解aaa3a11nnaaa313.1nn2nSaaan12naaaaaa)13124n33)11n3)1n.232S3＝．Sa＝1－n21nn3322S＝3n.3n2212且＝.當(dāng)N＋設(shè)a·Naaaa；＋n123n設(shè)b)NSbnSn＋nnnn12解令＝，121212＝.設(shè)Tannn12a，··n××3×，1212122121212121212××××，1T＝n112＋12＋12＋×，1212T＝22n1212T×n－nn即aaaa123n12，解＝12n1b))1nn2n＝，28b9b；nn當(dāng)9bn8或9Sn231a＋，nn213212n－n＝－a＝21.nn33n(2015·anSaaaSSNnn122n1.＋aa；2n求S.naSS，2n1NaSS＋11naaaa，2n1n1即a又aa2na，12aSSaaaa，3121121Naa.＋2na解aaa3a11nnaaa313.1nn2nSaaan12naaaaaa)13124n33)11n3)1n.232S3＝．Sa＝1－n21nn3322S＝3n.3n2212且＝.當(dāng)N＋設(shè)a·Naaaa；＋n123n設(shè)b)NSbnSn＋nnnn12解令＝，121212＝.設(shè)Tannn12a，··n××3×，1212122121212121212××××，1T＝n112＋12＋12＋×，1212T＝22n1212T×n－nn即aaaa123n12，解＝12n1b))1nn2n＝，28b9b；nn當(dāng)9bn8或9Sn231a＋，nn213212n－n＝－a＝21.nn33n(2015·anSaaaSSNnn122n1.＋aa；2n求S.naSS，2n1NaSS＋11naaaa，2n1n1即a又aa2na，12aSSaaaa，3121121Naa.＋2na解aaa3a11nnaaa313.1nn2nSaaan12naaaaaa)13124n33)11n3)1n.232S3＝．Sa＝1－n21nn3322S＝3n.3n2212且＝.當(dāng)N＋設(shè)a·Naaaa；＋