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2019-2020年高考数学一轮复习第六章数列第三节等比数列及其前n项和课后作业理一、选择题1在等比数列an中,如果a1a418,a2a312,那么这个数列的公比为()A2 B. C2或 D2或2(xx衡水模拟)已知正数组成的等比数列an,若a1a20100,那么a7a14的最小值为()A20 B25 C50 D不存在3(xx临沂模拟)已知等比数列an的前n项和为Sna2n1,则a的值为()AB.CD.4已知数列1,a1,a2,9是等差数列,数列1,b1,b2,b3,9是等比数列,则的值为()A. B. C. D.5已知Sn是等比数列an的前n项和,若存在mN*,满足9,则数列an的公比为()A2B2C3 D3二、填空题6(xx浙江高考)已知an是等差数列,公差d不为零若a2,a3,a7成等比数列,且2a1a21,则a1_,d_.7等比数列an满足an0,nN*,且a3a2n322n(n2),则当n1时,log2a1log2a2log2a2n1_.8在各项均为正数的等比数列an中,已知a2a416,a632,记bnanan1,则数列bn的前5项和S5为_三、解答题9已知an是首项为1,公差为2的等差数列,Sn表示an的前n项和(1)求an及Sn;(2)设bn是首项为2的等比数列,公比q满足q2(a41)qS40,求bn的通项公式及其前n项和Tn.10已知公比不为1的等比数列an的首项a1,前n项和为Sn,且a4S4,a5S5,a6S6成等差数列(1)求等比数列an的通项公式;(2)对nN*,在an与an1之间插入3n个数,使这3n2个数成等差数列,记插入的这3n个数的和为bn,求数列bn的前n项和Tn.1(xx兰州模拟)设Sn为数列an的前n项和,对任意的nN*,都有Snm1man(m为常数,且m0)(1)求证:数列an是等比数列;(2)设数列an的公比qf(m),数列bn满足b12a1,bnf(bn1)(n2,nN*),求数列bn的通项公式2设数列的前n项和为Sn,nN*.已知a11,a2,a3,且当n2时,4Sn25Sn8Sn1Sn1.(1)求a4的值;(2)证明:为等比数列;(3)求数列的通项公式答 案一、选择题1解析:选C设数列an的公比为q,由,得q2或q.2解析:选A(a7a14)2aa2a7a144a7a144a1a21400.a7a1420.3解析:选A当n2时,anSnSn1a2n1a2n2a2n2,当n1时,a1S1a,a,a.4解析:选C因为1,a1,a2,9是等差数列,所以a1a21910.又1,b1,b2,b3,9是等比数列,所以b199,易知b20,所以b23,所以.5解析:选B设公比为q,若q1,则2,与题中条件矛盾,故q1.qm19,qm8.qm8,m3,q38,q2.二、填空题6解析:a2,a3,a7成等比数列,aa2a7,(a12d)2(a1d)(a16d),即2d3a10.又2a1a21,3a1d1.由解得a1,d1.答案:17解析:由等比数列的性质,得a3a2n3a22n,从而得an2n.log2a1log2a2log2a2n1log2(a1a2n1)(a2a2n2)(an1an1)anlog22n(2n1)n(2n1)2n2n.答案:2n2n8解析:设数列an的公比为q,由aa2a416得,a34,即a1q24,又a6a1q532,解得a11,q2,所以ana1qn12n1,bnanan12n12n32n1,所以数列bn是首项为3,公比为2的等比数列,所以S593.答案:93三、解答题9解:(1)因为an是首项a11,公差d2的等差数列,所以ana1(n1)d2n1.故Snn2.(2)由(1)得a47,S416.因为q2(a41)qS40,即q28q160,所以(q4)20,从而q4.又因b12,bn是公比q4的等比数列,所以bnb1qn124n122n1.从而bn的前n项和Tn(4n1)10解:(1)因为a4S4,a5S5,a6S6成等差数列,所以a5S5a4S4a6S6a5S5,即2a63a5a40,所以2q23q10,因为q1,所以q,所以等比数列an的通项公式为an.(2)由题意得bn3nn,所以Tn.1解:(1)证明:当n1时,a1S1m1ma1,解得a11.当n2时,anSnSn1man1man,即(1m)anman1.又m为常数,且m0,(n2)数列an是首项为1,公比为的等比数列(2)由(1)得,qf(m),b12a12.bnf(bn1),1,即1(n2)数列是首项为,公差为1的等差数列(n1)1,即bn(nN*)2解:(1)当n2时,4S45S28S3S1,即45811,解得a4.(2)证明:由4Sn25Sn8Sn1Sn1(n2),得4Sn24Sn1SnSn14Sn14Sn(n2),即4an2an4an1(n2)4a3a14164a2,4an2an4an1对nN*都成立,数列是以a2a11为首项,为公比的等比数列(3)由(2)知,an1ann1,即4.数列是以2为首项,4为公差的等差数列,24(n1)4n2,即an(2n1)n1,数列的通项公式为an(2n1)n1.
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