高考數(shù)學(xué)數(shù)學(xué)文科數(shù)列專題練習(xí)

文科數(shù)學(xué)數(shù)列專題練習(xí)1.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}，a1a2a35,a7a8a910，則a4a5a6()A.52B.7C.6D.422.設(shè)Sn為等差數(shù)列{an}的前n項(xiàng)和，若a11，公差d2，Sk2Sk24，則k=()A.8B.7C.6D.53.已知數(shù)列{an}的前n項(xiàng)和為Sn，a11，Sn2an1，則Sn()A.2n1B.(3)n1C.(2)n1D.12432n14.已知數(shù)列{an}知足3an1an0，a2，則{an}的前10項(xiàng)和等于()31A.6(1310)B.(1310)C.3(1310)D.3(1310)95.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn.若S23,S415，則S6()A.31B.32C.63D.646.Sn是等差數(shù)列{an}的前n項(xiàng)和，若a1a3a53，則S5()A.5B.7C.9D.117.已知等比數(shù)列{an}知足a11,a3a54(a41)，則a2()411A.2B.1C.D.288.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前4項(xiàng)和為15，且a53a34a1，則a3()A.16B.8C.4D.29.記Sn為等差數(shù)列{an}的前n項(xiàng)和，若a35,a713，則S10.10.記等差數(shù)列{an}的前n項(xiàng)和為Sn，設(shè)S3=12，且2a1，a2，a3+1成等比數(shù)列，求Sn.11.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn，已知a2-=6，6a1+a3=30，求an和Sn.112.已知數(shù)列{an}中，a11n2，前n項(xiàng)和Snan.3(1)求a2，a3；(2)求{an}的通項(xiàng)公式.13.等差數(shù)列{an}中，a74,a192a9（I）求{an}的通項(xiàng)公式；(Ⅱ)設(shè)bn1，求數(shù)列{bn}的前n項(xiàng)和Snnan14.數(shù)列{an}知足a11,a22,an22an1an2.(1)設(shè)bnan1an，證明{bn}是等差數(shù)列；(2)求{an}的通項(xiàng)公式.215.已知各項(xiàng)都為正數(shù)的數(shù)列{an}知足a11，an2(2an11)an2an10.求a2，a3；求{an}的通項(xiàng)公式.16.設(shè)數(shù)列{an}知足a13a2...(2n1)an2n.求{an}的通項(xiàng)公式；求數(shù)列{an}的前n項(xiàng)和.2n117.等比數(shù)列{an}中，a11,a54a3.求{an}的通項(xiàng)公式；(2)記Sn為{an}的前n項(xiàng)和.若Sm63，求m.3廣西文科數(shù)學(xué)數(shù)列專題練習(xí)答案1.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}，a1a2a35,a7a8a910，則a4a5a6(A)A.52B.7C.6D.42【分析】a1a2a35,a235；同理a7a8a910a8310，a52a2a8a56(a2a8)3a23a8350應(yīng)選A。2.設(shè)Sn為等差數(shù)列{an}的前n項(xiàng)和，若a11，公差d2，Sk2Sk24，則k=(D)A.8B.7C.6D.5【分析】由等差求和公式Snna1n(n1)d得Snn2，則有：Sk2(k2)2,Skk2，2Sk2Sk24(k2)2k224解得k5。應(yīng)選D。3.已知數(shù)列{an}的前n項(xiàng)和為Sn，a11，Sn2an1，則Sn(B)A.2n1B.(3)n1C.(2)n1D.12132n1S1a12a2,a2時(shí)，Sn12anan2an2an，【分析】由題意得，當(dāng)n212即：an13，所以數(shù)列{an}(n2)是以a213an2為首項(xiàng)，q為公比的等比數(shù)列，221(1(3)n1)3n1Sn122,nN。應(yīng)選B。321244.已知數(shù)列{an}知足3an1an0，a2{an}的前10項(xiàng)和等于(C)，則13A.6(1310)B.(1310)C.3(1310)D.3(1310)9【分析】3anan0,an1111an，數(shù)列{an}是以4為首項(xiàng)，公比的等比數(shù)列，334[1(1)10]S1033(1310),。應(yīng)選C。1135.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn.若S23,S415，則S6(C)4A.31B.32C.63D.64【分析】S2a1a2,S4S2a3a4(a1a2)q2,S6S4a5a6(a1a2)q4，所以，S2,S4S2,S6S4是等比數(shù)列，即3，12，S615成等比數(shù)列，可得1223(S615)，解得，S663，應(yīng)選C。6.Sn是等差數(shù)列{an}的前n項(xiàng)和，若a1a3a53，則S5(A)A.5B.7C.9D.11【分析】數(shù)列{an}是等差數(shù)列，且a1a3a53得3a33，即a31.S55a35，應(yīng)選A.7.已知等比數(shù)列{an}知足a11,a3a54(a41)，則a2(C)411A.2B.1D.C.81211【分析】設(shè)等比數(shù)列{an}的公比為q，a1,a3a54(a41)，()2q64(q31)，444化為q38,解得q2，則a212142

.應(yīng)選C.8.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前4項(xiàng)和為15，且a53a34a1，則a3(C)A.16B.8C.4D.2【分析】設(shè)等比數(shù)列{an}的公比q（q>0），則由前4項(xiàng)和為15，且a53a34a1，有a1a1qa1q2a1q315a11a3224，應(yīng)選C.a1q43a1q24a1，，q29.記Sn為等差數(shù)列{an}的前n項(xiàng)和，若a35,a713，則S10100.【分析】在等差數(shù)列{an}中，由a35,a713，得da7a31357342，1011092a1a32d541.則S10100.故答案為：100.210.記等差數(shù)列{an}的前n項(xiàng)和為Sn，設(shè)S3=12，且2a1，a2，a3+1成等比數(shù)列，求Sn.解：設(shè)等差數(shù)列{an}的公差為d，由題意得a222a1(a31)a11a183a132解得或d4d12d32Sn1n(3n1)或Sn2n(5n).211.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn，已知a2-=6，6a1+a3=30，求an和Sn.5解：設(shè)等比數(shù)列{an}的公比為q，由題意得：a1q6a13a12，解得或.6a1a1q230q2q3當(dāng)a13,q2時(shí)：an32n1,Sn3(2n1)；當(dāng)a12,q3時(shí)：an23n1,Sn3n1.12.已知數(shù)列{an}中，a11，前n項(xiàng)和Snn2an.3求a2，a3；求{an}的通項(xiàng)公式.解：（1）數(shù)列{an}中，由a11，前n項(xiàng)和Snn2an可得：S24a2,3(a1a2)4a2，33解得a23a13，由S35a3得3(a1a2a3)5a3，解得a33a2)6.3(a12（2）由題意知a11，當(dāng)n1時(shí)，有anSnn2n1Sn1an3an1，整理得：n1an3a14a23nn1an1，an1，于是a11，a2，a3，...，an1nan2，ann1122n1將以上n理得：ann(n1)個(gè)式子兩頭分別相乘，整2.綜上，{an}的通項(xiàng)公式為n(n1)an.213.等差數(shù)列{an}中，a74,a192a9（II）求{an}的通項(xiàng)公式；(Ⅱ)設(shè)bn1，求數(shù)列{bn}的前n項(xiàng)和Snnan解：（Ⅰ）設(shè)等差數(shù)列{an}的公差為d，a74,a192a9，a16d4解a118d2(a18d)得，a11,d1an1(n1n211).22（Ⅱ）bn1222，Sn2(1111...11)nann(n1)nn1223nn12(1n1)2n1n114.數(shù)列{an}知足a11,a22,an22an1an2.6(1)設(shè)bnan1an，證明{bn}是等差數(shù)列；求{an}的通項(xiàng)公式.解：（1）由an22an1an2得：an2an1an1an2，由bnan1an得：bn1bn2，即bn1bn2，又b1a2a11，{bn}是以首項(xiàng)為1，公差為2的等差數(shù)列.（2）由（1）得，bn12(n1)2n1，由由bnan1an得：an1an2n1，則a2a11,a3a23,a4a35,...,anan12(n1)1，所以，ana1135...2(n1)1(n1)(12n3)(n1)2,又a11，所2以{an}的通項(xiàng)公式an(n1)21n22n2.15.已知各項(xiàng)都為正數(shù)的數(shù)列{an}知足a11，an2(2an11)an2an10.求a2，a3；求{an}的通項(xiàng)公式.解：（1）依照題意，an2(2an11)an2an10，當(dāng)n1時(shí)，有a12(2a21)a12a20,而a1，則有1(2a1)a2a0，解得a21，同理，當(dāng)n2時(shí)，有121221111a22(2a31)a22a30,又由a2，解得a3，故a2,a3.2424（2）由題意得，an2(2an11)an2an10(an2an1)(an1)0，即有an2an1或an1，又由數(shù)列{an}各項(xiàng)都為正數(shù)，則有an2an1，故數(shù)列{an}是首項(xiàng)為a11，1的等比數(shù)列，則an1(1n11公比為2)n1.2216.設(shè)數(shù)列{an}知足a13a2...(2n1)an2n.求{an}的通項(xiàng)公式；求數(shù)列{an}的前n項(xiàng)和.2n1解：（1）數(shù)列{an}知足a13a2...(2n1)an2n.7當(dāng)n2時(shí)，a13a2...(2n3)a2時(shí)，a12上式也建立.an2n1

n12(n1).(2n1)an2an2.當(dāng)n12n1.（2）由題意得an211，2n1(2n1)(2n1)2n12n1數(shù)列{an}的前n項(xiàng)和=(11)(11)...(111)1112n.2n13352n2n12n2n117.等比數(shù)列{an}中，a11,a54a3.求{an}的通項(xiàng)公式；(2)記Sn為{an}的前n項(xiàng)和.若Sm63