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1、最新北师大版数学精品教学资料第一章3第4课时一、选择题1已知数列an满足3an1an0,a2,则an的前10项和等于()A6(1310)B(1310)C3(1310)D3(1310)答案C解析本题考查等比数列的定义,前n项和的求法3an1an0qa2a1qa1,a14S103(1310)2设Sn为等比数列an的前n项和,已知3S3a42,3S2a32,则公比q()A3B4C5D6答案B解析3S3a42,3S2a32,3S33S2a4a3,3a3a4a3,4a3a4,4,q4.3若等比数列an满足anan164n,则公比为()A2B4C8D16答案C解析本题考查了灵活利用数列的特点来解题的能力a
2、nan164n,an1an64n1q264q8.4在各项为正数的等比数列中,若a5a4576,a2a19,则a1a2a3a4a5的值是()A1061B1023C1024D268答案B解析由题意得a4(q1)576,a1(q1)9,q364,q4,a13,a1a2a3a4a51023.5在等比数列an中,a11,公比|q|1,若ama1a2a3a4a5,则m()A9B10C11D12答案C解析a11,ama1a2a3a4a5aq10q10,又ama1qm1qm1,qm1q10,m110,m11.6已知等比数列前20项和是21,前30项和是49,则前10项和是()A7B9C63D7或63答案D解析
3、由S10,S20S10,S30S20成等比数列,(S20S10)2S10(S30S20),即(21S10)2S10(4921),S107或63.二、填空题7已知数列an中,an,则a9_.设数列an的前n项和为Sn,则S9_.答案256377解析a928256,S92022242628371115377.8在等比数列an中,已知对于任意nN,有a1a2an2n1,则aaa_.答案4n解析a1a2an2n1,a1a2an12n11(n2),两式相减,得an2n12n112n2n12n1,a(2n1)222n24n1,aaa4n.三、解答题9(2014北京文,15)已知an是等差数列,满足a13,
4、a412,数列bn满足b14,b420,且bnan为等比数列(1)求数列an和bn的通项公式;(2)求数列bn的前n项和解析(1)设等差数列an的公差为d,由题意得d3.所以ana1(n1)d3n(n1,2,)设等比数列bnan的公比为q,由题意得q38,解得q2.所以bnan(b1a1)qn12n1,从而bn3n2n1(n1,2,)(2)由(1)知bn3n2n1(n1,2,)数列3n的前n项和为n(n1),数列2n1的前n项和为12n1.所以,数列bn的前n项和为n(n1)2n1.10求和Sn12422723(3n2)2n.解析Sn124227233(n1)22n1(3n2)2n2Sn122
5、4233(n1)22n(3n2)2n1得,Sn1232232332n(3n2)2n13(2222n)(3n2)2n143(2n12)(3n2)2n1432n163n2n12n242n23(1n)2n110.Sn3(n1)2n12n210(3n5)2n110.一、选择题1已知等比数列an中,公比q,且a1a3a5a9960,则a1a2a3a100()A100B90C120D30答案B解析a2a4a6a100a1qa3qa5qa99qq(a1a3a5a99)6030a1a2a3a100(a1a3a5a99)(a2a4a6a100)603090.2数列an的前n项和为Sn,若a11,an13Sn(n
6、1),则a6()A344B3441C45D451答案A解析该题考查已知一个数列的前n项和Sn与an1的关系,求通项公式an.注意的问题是用anSnSn1时(n2)的条件an13Snan3Sn1得an1an3Sn3Sn13an即an14an4.(n2)当n2时,a23a13,34an为从第2项起的等比数列,且公比q4,a6a2q4344.3设an是任意等比数列,它的前n项和、前2n项和与前3n项和分别为X、Y、Z,则下列等式中恒成立的是()AXZ2YBY(YX)Z(ZX)CY2XZDY(YX)X(ZX)答案D解析由题意知SnX,S2nY,S3nZ.又an是等比数列,Sn,S2nSn,S3nS2n
7、为等比数列,即X,YX,ZY为等比数列,(YX)2X(ZY),整理得Y2XYZXX2,即Y(YX)X(ZX)故选D4设等比数列an的前n项和为Sn,若3,则()A2BCD3答案B解析3,S63S3,2,S3,S6S3,S9S6成等比,22,S94S3S67S3,选B二、填空题5等比数列an的前n项和为Sn3n1m,则a1_.答案6解析a1S19m,a2S2S127m9m18,a3S3S281m27m54,又an为等比数列,aa1a3,18254(9m),解得m3.a19m6.6(2014天津理,11)设an是首项为a1,公差为1的等差数列,Sn为其前n项和,若S1,S2,S4成等比数列,则a1
8、的值为_答案解析本题考查等差数列等比数列综合应用,由条件:S1a1,S2a1a2a1a1d2a11S4a1a2a3a4a1a1da12da13d4a16d4a16(2a11)2a1(4a16)4a14a14a6a1a1.三、解答题7已知数列an和bn中,数列an的前n项和为Sn.若点(n,Sn)在函数yx24x的图像上,点(n,bn)在函数y2x的图像上(1)求数列an的通项公式;(2)求数列anbn的前n项和Tn.解析(1)由已知得Snn24n,当n2时,anSnSn12n5,又当n1时,a1S13,符合上式an2n5.(2)由已知得bn2n,anbn(2n5)2n.Tn321122(1)2
9、3(2n5)2n,2Tn322123(2n7)2n(2n5)2n1.两式相减可得Tn6(23242n1)(2n5)2n1(2n5)2n16(72n)2n114.8设数列an满足a12,an1an322n1.(1)求数列an的通项公式;(2)令bnnan,求数列bn的前n项和Sn.解析(1)由已知,当n1时,an1(an1an)(anan1)(a2a1)a13(22n122n32)222n122(n1)1.而a12,符合上式,所以数列an的通项公式为an22n1.(2)由bnnann22n1,知Sn12223325n22n1,22Sn123225327(n1)22n1n22n1.,得(122)Sn2232522n1n22n1,即Sn(3n1)22n12