數(shù)列求和方法大全例題變式解析答案-強烈推薦

,數(shù)列前n項和求法aaaan1n12"ab}a}b}qnnnnqSS和nn"12111241，1111242n1-1357(1)(2n1)SSn1nn`111aaaanaan1n1223an、aaaSn12n|CC(2nC(nC012nnnnnnsin2sin3sin89sin12222|12x3x2nx(x0)Sn1n：a,5a,,(2na(a0)n2n1a,2a,3a,,na,n32n123n求和：Saaa3ann2^1111111，1，11224242n1；5n92：13,24,35,,n(n2),；#n2（）2nnn.1357(1)(2n1)Sn1nS…（-14）nNnn1}已知數(shù)列{aa=2，a+a=1，S為{a}前n項和，求Snn1nn+1nn【已知數(shù)列{aa=1，a=4，a=a+2為{a}前n項和，求n-2n12nnnSn、1111aaaaaaadSn1aan1nn1223341111,,,,,；132435n(n!1annn1n2242(2n)2(2nnS1335n-daan1123；n123n;aaa36,a}annS.n9SSnnnaaaT.n12na2an1p,q使數(shù)列變式2：設(shè)數(shù)列a}滿足an，已知存在常數(shù)1n1napn}aaa.n12n"1ababnS.a212nnnnn:、Saaan12n|a(ad)[a(nd]①111Sa(ad)[a(n1)d]②nnnn個2Saa(aa)(aa)n1n1n1nn(aa)1nn(aa)S1n2n|CCmnmnnCCC(2nCS012nnnnnn(2nC(2nCCCC(2)則Snn1210nnnnnnCCmnmnn(2)有:2S(2n2)C(2n2)C(2n2)C(2n2)C012nnnnnnS(nCCCC](n2012nnnnnnnsin1sin2sin3sin89，設(shè)S2222|sin89sin88sin87sin1，S22222∴2S89，S．S12x3xnx(x0)2n1nn(n1)S=n212x3xnxS2n1nxSx2x3xnx23nn【(1x)S1xxxnn12n1xnnxn1x1xnnxnS∴x)1xn2,aa,a,a,20n11解：S13a5a(2na2n1nnaa5a(2na223n12:a)S12a2a2a2a(2na23n1nn"2aa)n1當a,(1a)S1(2nna)2n1a(2na(2nann1Sa)2n,Sn當a2nSa2aa，23nnn(n2當a1S123…n，n1S…na，a2aa當a23nna2aa…na，234n1na(1a)na)Saaa，n1…ann1231an。(naan2n1∴S．a(chǎn))2n12S3naaa3ann2n(n時，a1S123nn2a時，因為a0⑵12Saa3n①aann23|112n1nS②aaaaan1n23n1111n)Saaaaan1n2n11)naan1an11aa(an(an所以Sa(ann2綜上所述，n(n(aa2Sa(an(anna(an21，1121111111，24242n1111a1242n1n11()n12212n1121111))∴S224n~11124)2n111(21)(2)(2)2221(2)2n1112n241)2n112n22n1個S55個(999999999)59n523n9)55n]n．23nn992)n2n∵n(n，2n(nn(123n)2(123∴……n)．22226<n=(kN+SS3)(57)n2k[(4k(4k2kn當n2kkN時，SSSa2k[(4k1)]n2k12k2k2k1n](1)nSn1nn解：當n為偶數(shù)時：S15913nn4nn24當n為奇數(shù)時：S513-32nSaaaa…aa解：①當n為偶數(shù)時：n1234n1nnn(aa)(aa)…(aa)1n1n221234Saaa)aa)aa)②當n為奇數(shù)時：n12345n1n?n1n3222解：∵a-a=2（n≥3）nn-2∴a,a,a,…,a為等差數(shù)列；a,a,a,…,a為等差數(shù)列1352n-12462nn1當n為奇數(shù)時：a1(1)?2n2nn當n為偶數(shù)時：a4(1)?2n22n即n∈N時，+an1(1)nnn1n(n1)S23…n)2n1∴①n為奇數(shù)時：22n!nn1)S(123…)2n②n為偶數(shù)時：22n111ada∵kkaaka(ad)da(ad)k1kkkk11(daad1111)(daa)kkkk1111111()()∴Sdaa1daa2n23~11(da1)an1n111)()(daaaa1111aa1223n1n111()daa1nn1a[a(nd]11111(1∵)，n(n2nn2111111111111)()()(∴S)．32435nn222n1n2n2】1n1nn1(nn1nann1)(n1n)n111Sn∴2132n1n(2(3(n1n)n11．解:(2k)2(2k)111111(12a1)(2kk(2kk(2kk22k12k1k1111111n212n(n2n1Saaan)()()23352n12n12n1n12n(nn(a123n2SSa(an(anaaaann(a23na(an2a，2132即d2或a或a.**nnnSnn當|n123n1n＋2221當＝n－222123nn1－n＋222n1231n－222n或21時，20Sn31232nn,n2122