高三數(shù)列知識點與題型總結(jié)(文科)說課講解

數(shù)列考點總結(jié)第一部分求數(shù)列的通項公式一、數(shù)列的相關(guān)概念與表示方法（見輔導書）二、求數(shù)列的通項公式aaf(n)n1naaf(n)(n若n1n，aaf21aaf32aaf(n)則n1nnaa()fnn11k11已知數(shù)列a}nn1aa2n，a1，求數(shù)列a}nn12已知數(shù)列a}nn1aa23，a3，求數(shù)列a}n1nn2n(nN)a*aaan1n.nn21nn1aa(n2)a}3a1n(nnn練習已知數(shù)列n滿足，，求此數(shù)列的通項公式.11a2nnaaaaf(n),n1nn1a項n.n;n;n;n1nS(a)例已知數(shù)列a}n,a0n且2n.a}annn2aan1,a1n3a}n2，求數(shù)列a}na1naf(n)an1naaaaf(n)f，f，，f(n)n123n1aaaa若1n2nanaf(k)n1a11k1例4已知數(shù)列a}nn1a2(n1)5a，a3，求數(shù)列a}n1nnn1a2na2aa0a例設(shè)1n1nn1nn2（，，3nanaqaf(n)n1nacad,(c0aan1n型1{{c且d0an;an;{an.(aca)n1n,aca(cacad,得n1nn1nddd,(c0)ac(a)(cdc1c1c1nn1ddaa11ncc1cdddda(a)cn1a(a)cn1c1c1c1c1n1n1.dddac(a)a}acadc1c1c1n1nnn1ncddacn1(a)1cc1n11acadacadn1nn有nn1,aac(aa)aa}aac(aa)nn1nnn1cn1n.n1n21,.a}a1,a2an2)a例nn1nn1apaqnp1apaqqnn1naaqn.n1nn，n1npn111papaqpa()pqbnbb()n1n1nnnnnpn1npqn令.qn1.1an1qn1paqqnnq,1apbnbbnqn1qnqn令5aq1pap)(nn設(shè).n1np例a}nn1a2a43，a1n1n1anaa12,a＝2,nNn1a1a*a}nn2’2.baab{}令nn1nna}求n12521a()(nN)n1*b332n）1）。naaf(n)n1n型af(n)n1a型naaf(n)型n1naadad{nn1n;aaf(n)為n兩n1naaf(n)f(n得.n1n1aaf(n)n1n型aapn1n{an;aaf(n為nnn1.0aa2n例{an1,n1n.1aaa(),(nN)n*a滿足21nn1例n.第二部分數(shù)列求和nq1或n＝；2n；2n.2nnnnnSn)nnnn122nn22SnanSn)nnnACBDaa)1k1kABDCannnnn1111，，nan123123362498anbbaabnS.nnnnnnn..例anSkaaa.nnn263求a；nnT.nnanSS3.nnnn求kanabnabbnT.＋2nnnnn141nT＝－－.nn1＋n22·33例anSaSN．*nn1nnan2設(shè)b＝bnT.aann1＋nnnaaNaaaa*n13215an111SSS1設(shè)babnS＋＋<kN*n4nnnSn123kanSnanT()2nnnnA2n2nn22naa2且aa22San2012n1n1n1nn2aaaaa2是aan234324an1若baaSbbbS.2nnnn12nnaaaaan1139ananS.nnSn1＋n31x