第6讲 数列求通项(累加法)(解析版)

第6讲f(n)(nN*)aaaaann1nnna(aa)(aa)(aa)aff(2)f(3)f(n(n2)a求n12132nn11aaf(1)21aaf(2)32aaf(3)43aaf(n1)nn1n1(aa)(aa)(aa)(aa)f(1)f(2)f(3)f(n1)=213243nn1aaf(1)fff(n=n13,aa2n2aa.n1n1nn1an21nann2a21n1nn2n1n2a2n2aaan1nn1nn1aaaaaaa222当an212n1213243nn1212n2n12a31aa，n以，21当时n1；21aann12nnn111,a1a(n1)2nnaa．nn1n1annbbnn21nNbn*nanan1nb2bn1b2nn1nnnaa2，nnn1n1nbb2a1，nbn1n112,......,bb2，n2b2,bb当bn2n11nn1n1n221bb22n21，2n12n221bn11n1n112nb；21n12n1当bn1n2121nNb；n*nn1n1题aa2n22题nn1n1an1nannbb2nn1naa3,aa23nN．nn1n1nan；a3nnana23，a23naan1n1nnn1aa23，，aa23，n2n1n221631n1，aa232323332n1n31n1a33a1nnaa3nnnaaaa2n.n1n1nan；a2nna，a2nn1naaaaaaa当nan12132n1212n1222222n2n112又a2a2n1n11，，，，a与610153nan1nN,n2a*nn(n1)ana，a；n2nn1ana1，n2aan1nn1n(n1)aaaaaaaan(n1)21nnn1n1n22112n(n1)故an2a34aa2，2a．nn1n1nan2n1；an3aa4nn1n23a，a4nn1n23，aa412213aa4，22323aa432433aa4，nn1n2334)naa44442(41)，123nn14n1122又a，22n1a2124nn122n1a2n1)1n3nan1na，.a1an2nn12nann；an2nnaa1n12nnnn1nnaa1n1.aann1nn1nn12n2na22a111a33a221aa1，，nn122nn12n231n11aa111111211，n22232n2212n2n12n.ann2n13当，n1a1122n.2nann(nanN*)a1.anan1n1nan2n3an21.S3nn2naan111n(n1)an1n1nn(nnn1naa11nan(n(1nn1n1nnn1n1nana11aa1，，，121n1nn1n1n212aa11，a2nn，1nn1nn又an；21n=1()nn项和，若出现bbfnnf(n)nn1bf(n)n(n，nbn1aan111n1n1nn(nnn1aa1，n1ana2n1n1n1nann;an设cn，cca1c2n1cn2n11n1nn1nan1nan21n设cn，c2n1ca1cn