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1、第45練 數(shù)列的遞推關(guān)系及通項(xiàng) 基礎(chǔ)保分練1已知數(shù)列an的通項(xiàng)公式為an則a20a15_.2在數(shù)列an中,a12,an12an3,則數(shù)列an的通項(xiàng)公式an_.3已知數(shù)列an滿足a10,an1(n1,2,3,),則a2019_.4已知數(shù)列an的前n項(xiàng)和為Sn,且Sn2an2,則a2019_.5由a11,an1給出的數(shù)列an的第34項(xiàng)是_6正項(xiàng)數(shù)列an中,滿足a11,a2,(nN*),那么an_.7已知數(shù)列an的首項(xiàng)a12,且(n1)annan1,則a5_.8數(shù)列an中,a11,且an1an2n,則a9_.9數(shù)列an中,若a11,an1an,則an_.10已知數(shù)列an的前n項(xiàng)和為Sn,a11,2S
2、n(n1)an,則an_.能力提升練1已知數(shù)列an的通項(xiàng)為an,當(dāng)an取得最小值時(shí),n的值為_2已知數(shù)列an,若a12,an1an2n1,則a2019_.3設(shè)各項(xiàng)均為正數(shù)的數(shù)列an的前n項(xiàng)和為Sn,且Sn滿足2S(3n2n4)Sn2(3n2n)0,nN*,則數(shù)列an的通項(xiàng)公式是_4已知數(shù)列an中,a12,n(an1an)an1,nN*.若對(duì)于任意的t0,1,nN*,不等式an1,則實(shí)數(shù)a的取值范圍是_6設(shè)數(shù)列an的前n項(xiàng)和為Sn,且Snn2n1,正項(xiàng)等比數(shù)列bn的前n項(xiàng)和為Tn,且b2a2,b4a5,數(shù)列cn中,c1a1,且cncn1Tn,則cn的通項(xiàng)公式為_答案精析基礎(chǔ)保分練11652.(2
3、)n113.4.220195.6.解析由已知2,數(shù)列是等比數(shù)列,又1,2,q2,2n1,an.710解析由題意得an1an,a2a14,a3a26,a4a38,a5a410.8511解析由題意可得an1an2n,則a9a1(a2a1)(a3a2)(a9a8)1212228291511.9.解析a11,an1an,則(n1)an1nana11,an.10n解析由2Sn(n1)an知,當(dāng)n2時(shí),2Sn1nan1,得2an(n1)annan1,(n1)annan1,當(dāng)n2時(shí),1,ann.能力提升練1152.20203.an3n24(,13,)解析根據(jù)題意,數(shù)列an中,n(an1an)an1,nan1(n1)an1,a1,233,2t2(a1)ta2a3恒成立,32t2(a1)ta2a3.2t2(a1)ta2a0,在t0,1上恒成立,設(shè)f(t)2t2(a1)ta2a,t0,1,即解得a1或a3.5.6cn2nn解析Snn2n1,令n1,a11,anSnSn12(n1)(n2),經(jīng)檢驗(yàn)a11不符合上式,an又?jǐn)?shù)列bn為等比數(shù)列,b2a22,b4a58,q24,又?jǐn)?shù)列bn為正項(xiàng)等比數(shù)列,q2,b11,bn2n1.Tn2n1,c2c1211,c3c2221,cncn12n11,以上各式相加得cnc1(n1),c1a11,cn12nn1,cn2nn.4