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1、(全國(guó)通用版)2022年高考數(shù)學(xué)一輪復(fù)習(xí) 第八單元 數(shù)列雙基過(guò)關(guān)檢測(cè) 理一、選擇題1(2017全國(guó)卷)記Sn為等差數(shù)列an的前n項(xiàng)和若a4a524,S648,則an的公差為()A1B2C4 D8解析:選C設(shè)等差數(shù)列an的公差為d,由得即解得d4.2(2018江西六校聯(lián)考)在等比數(shù)列an中,若a3a5a73,則a2a8()A3 B.C9 D13解析:選A由a3a5a73,得a3,即a5,故a2a8a3.3在數(shù)列an中,已知a12,a27,an2等于anan1(nN*)的個(gè)位數(shù),則a2 018()A8 B6C4 D2解析:選D由題意得a34,a48,a52,a66,a72,a82,a94,a108
2、.所以數(shù)列中的項(xiàng)從第3項(xiàng)開始呈周期性出現(xiàn),周期為6,故a2 018a33568a82.4已知數(shù)列an滿足a11,anan12n(n2,nN*),則a7()A53 B54C55 D109解析:選Ca2a122,a3a223,a7a627,各式相加得a7a12(2347)55.5設(shè)數(shù)列an的前n項(xiàng)和為Sn,若a11,an13Sn(nN*),則S6()A44 B45C.(461) D.(451)解析:選B由an13Sn,得a23S13.當(dāng)n2時(shí),an3Sn1,則an1an3an,n2,即an14an,n2,則數(shù)列an從第二項(xiàng)起構(gòu)成等比數(shù)列,所以S645.6等差數(shù)列an和bn的前n項(xiàng)和分別為Sn,Tn
3、,對(duì)一切自然數(shù)n,都有,則等于()A. B.C. D.解析:選CS99a5,T99b5,.7已知數(shù)列an是首項(xiàng)為1的等比數(shù)列,Sn是其前n項(xiàng)和,若5S2S4,則log4a3的值為()A1 B2C0或1 D0或2解析:選C由題意得,等比數(shù)列an中,5S2S4,a11,所以5(a1a2)a1a2a3a4,即5(1q)1qq2q3,q3q24q40,即(q1)(q24)0,解得q1或2,當(dāng)q1時(shí),a31,log4a30.當(dāng)q2時(shí),a34,log4a31.綜上所述,log4a3的值為0或1.8設(shè)數(shù)列an是公差為d(d0)的等差數(shù)列,若a1a2a315,a1a2a380,則a11a12a13()A75
4、B90C105 D120解析:選C由a1a2a315得3a215,解得a25,由a1a2a380,得(a2d)a2(a2d)80,將a25代入,得d3(d3舍去),從而a11a12a133a123(a210d)3(530)105.二、填空題9若數(shù)列an滿足a13a232a33n1an,則數(shù)列an的通項(xiàng)公式為_解析:當(dāng)n2時(shí),由a13a232a33n1an,得a13a232a33n2an1,兩式相減得3n1an,則an.當(dāng)n1時(shí),a1滿足an,所以an.答案:an10數(shù)列an的前n項(xiàng)和為Sn,若Sn2an1,則an_.解析:Sn2an1, Sn12an11(n2), 得an2an2an1,即an
5、2an1.S1a12a11,即a11,數(shù)列an為首項(xiàng)是1,公比是2的等比數(shù)列,故an2n1.答案:2n111已知數(shù)列an中,a2na2n1(1)n,a2n1a2nn,a11,則a20_.解析:由a2na2n1(1)n,得a2na2n1(1)n,由a2n1a2nn,得a2n1a2nn,故a2a11,a4a31,a6a51,a20a191.a3a21,a5a42,a7a63,a19a189.又a11,累加得:a2046.答案:4612數(shù)列an為正項(xiàng)等比數(shù)列,若a33,且an12an3an1(n2,nN*),則此數(shù)列的前5項(xiàng)和S5_.解析:設(shè)公比為q(q0),由an12an3an1,可得q22q3,
6、所以q3,又a33,則a1,所以此數(shù)列的前5項(xiàng)和S5.答案:三、解答題13已知在等差數(shù)列an中,a35,a1a1918.(1)求公差d及通項(xiàng)an;(2)求數(shù)列an的前n項(xiàng)和Sn及使得Sn取得最大值時(shí)n的值解:(1)a35,a1a1918,an112n.(2)由(1)知,Snn210n(n5)225,n5時(shí),Sn取得最大值14已知數(shù)列an滿足n2n.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn,求數(shù)列bn的前n項(xiàng)和Sn.解:(1)n2n,當(dāng)n2時(shí),(n1)2n1,兩式相減得2n(n2),ann2n1(n2)又當(dāng)n1時(shí),11,a14,滿足ann2n1.ann2n1.(2)bnn(2)n,Sn1(2)12(2)23(2)3n(2)n.2Sn1(2)22(2)33(2)4(n1)(2)nn(2)n1,兩式相減得3Sn(2)(2)2(2)3(2)4(2)nn(2)n1n(2)n1n(2)n1,Sn.