1、高中数学必修五等比数列的前n项和等比数列的前n项和第1课时一、选择题1设等比数列an的前n项和Sn,已知a12,a24,那么S10等于()A2102 B292C2102 D2112答案D解析q2,S102(2101)2112,选D2等比数列an的前n项和Sn3na,则a的值为()A3 B0C1 D任意实数答案C解析S1a13a,S2S1a232a3a6,S3S2a333a32a18,所以a1.3设数列an是等比数列,其前n项和为Sn,且S33a3,则公比q的值为()A BC1或 D1或答案C解析当q1时,S33a13a3符合题意;当q1时,S33a1q2.a10,1q33q2(1q)由1q0,
2、两边同时约去1q,得1qq23q2,即2q2q10,解得q.综上,公比q1,或q.4已知an是等比数列,a22,a5,则a1a2a2a3anan1()A16(14n) B16(12n)C(14n) D(12n)答案C解析q3,q.anan14()n14()n252n,故a1a2a2a3a3a4anan123212123252n(14n)5(2014大纲全国卷文,8)设等比数列an的前n项和为Sn.若S23,S415,则S6()A31 B32C63 D64答案C解析解法1:由条件知:an0,且q2.a11,S663.解法2:由题意知,S2,S4S2,S6S4成等比数列,即(S4S2)2S2(S6
3、S4),即1223(S615),S663.6已知等比数列前20项和是21,前30项和是49,则前10项和是()A7 B9C63 D7或63答案D解析由S10,S20S10,S30S20成等比数列,(S20S10)2S10(S30S20),即(21S10)2S10(4921),S107或63.二、填空题7设an是公比为正数的等比数列,若a11,a516,则数列an的前7项和为_答案127解析设数列an的公比为q(q0),则有a5a1q416,q2,数列的前7项和为S7127.8已知Sn为等比数列an的前n项和,Sn93,an48,公比q2,则项数n_.答案5解析由Sn93,an48,公比q2,得
4、2n32n5.三、解答题9已知an是公差不为零的等差数列,a11,且a1,a3,a9成等比数列(1)求数列an的通项公式; (2)求数列2an的前n项和Sn.解析(1)由题设,知公差d0,由a11,a1,a3,a9成等比数列得,解得d1,或d0(舍去)故an的通项an1(n1)1n.(2)由(1)知2an2n,由等比数列前n项和公式,得Sn222232n2n12.10等比数列an的前n项和为Sn,已知S1,S3,S2成等差数列(1)求an的公比q;(2)若a1a33,求Sn.解析(1)S1,S3,S2成等差数列,2S3S1S2,q1不满足题意a1,解得q.(2)由(1)知q,又a1a3a1a1
5、q2a13,a14.Sn1()n.一、选择题1若等比数列an各项都是正数,a13,a1a2a321,则a3a4a5的值为()A21 B42C63 D84答案D解析a1a2a321,a1(1qq2)21,又a13,1qq27,an0,q0,q2,a3a4a5q2(a1a2a3)222184.2等比数列an中,已知前4项之和为1,前8项和为17,则此等比数列的公比q为()A2 B2C2或2 D2或1答案C解析S41,S8S4q4S41q417q2.3在各项为正数的等比数列中,若a5a4576,a2a19,则a1a2a3a4a5的值是()A1 061 B1 023C1 024 D268答案B解析由a
6、4(q1)576,a1(q1)9,q364,q4,a13,a1a2a3a4a51 023.4设an是由正数组成的等比数列,Sn为其前n项和,已知a2a41,S37,则S5()A BC D答案B解析an是正数组成的等比数列,a31,又S37,消去a1得,7,解之得q,a14,S5.二、填空题5设等比数列an的前n项和为Sn,若a11,S64S3,则a4_.答案3解析若q1时,S33a1,S66a1,显然S64S3,故q1,4,1q34,q33.a4a1q33.6已知等比数列an共有2n项,其和为240,且奇数项的和比偶数项的和大80,则公比q_.答案2解析由题意,得,解得S奇80,S偶160,q
7、2.三、解答题7已知各项均为正数的等比数列an的前n项和为Sn,S3,首项a1.(1)求数列an的通项公式an;(2)令bn6n61log2an,求数列bn的前n项和Tn.解析(1)由已知S3a1a2a3,qq2.q2q60,(q3)(q2)0q2或q3.(舍)ana1qn12n2.(2)bn6n61log22n26n61n27n63.bnbn17n637n7637,数列an是等差数列又b156,Tnnb1n(n1)756nn(n1)7n2n.8(2014北京文,15)已知an是等差数列,满足a13,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;
8、(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.第2课时一、选择题1数列1,3,5,7,的前n项和Sn为()An21 Bn21Cn22 Dn22答案A解析由题设知,数列的通项为an2n1,显然数列的各项为等差数列2n1和等比数列相应项的和,从而Sn13(
9、2n1)()n21.2已知数列an的通项公式是an,若前n项和为10,则项数n为()A11 B99C120 D121答案C解析因为an,所以Sna1a2an(1)()()110,解得n120.3已知等比数列的前n项和Sn4na,则a的值等于()A4 B1C0 D1答案B解析a1S14a,a2S2S142a4a12,a3S3S243a42a48,由已知得aa1a3,14448(4a),a1.4数列an的通项公式为an(1)n1(4n3),则它的前100项之和S100等于()A200 B200C400 D400答案B解析S10015913(4993)(41003)50(4)200.5数列an的前n
10、项和为Sn,若an,则S5等于()A1 BC D答案B解析an,S511.6数列an中,已知对任意nN*,a1a2a3an3n1,则aaaa等于()A(3n1)2 B(9n1)C9n1 D(3n1)答案B解析a1a2a3an3n1,a1a2a3an13n11(n2),两式相减得an3n3n123n1,又a12满足上式,an23n1.a432n249n1,aaa4(19929n1)(9n1)二、填空题7数列,前n项的和为_答案4解析设Sn Sn 得(1)Sn2.Sn4.8已知数列a12,a24,ak2k,a1020共有10项,其和为240,则a1a2aka10_.答案130解析由题意,得a1a2
11、aka10240(242k20)240110130.三、解答题9求数列1,3a,5a2,7a3,(2n1)an1的前n项和解析当a1时,数列变为1,3,5,7,(2n1),则Snn2,当a1时,有Sn13a5a27a3(2n1)an1, aSna3a25a37a4(2n1)an, 得:SnaSn12a2a22a32an1(2n1)an,(1a)Sn1(2n1)an2(aa2a3a4an1)1(2n1)an21(2n1)an.又1a0,所以Sn.10(2014全国大纲文,17)数列an满足a11,a22,an22an1an2.(1)设bnan1an,证明bn是等差数列;(2)求an的通项公式解析
12、(1)证明:由an22an1an2得an2an1an1an2.即bn1bn2.又b1a2a11.所以bn是首项为1,公差为2的等差数列(2)由(1)得bn12(n1)2n1,即an1an2n1.于是(ak1ak)(2k1),所以an1a1n2,即an1n2a1.又a11,所以an的通项公式为ann22n2.一、选择题1已知等差数列an和bn的前n项和分别为Sn,Tn,且,则()A BC6 D7答案A解析,又,.2数列an满足an1(1)nan2n1,则an的前60项和为()A3690 B3660C1845 D1830答案D解析不妨令a11,则a22,a3a5a71,a46,a610,所以当n为
13、奇数时,an1;当n为偶数时,各项构成以2为首项,4为公差的等差数列,所以前60项的和为3023041830.3数列an的通项公式是ansin(),设其前n项和为Sn,则S12的值为()A0 BC D1答案A解析a1sin()1,a2sin()1,a3sin()1,a4sin(2)1,同理,a51,a61,a71,a81,a91,a101,a111,a121,S120.4已知等差数列an满足a5a2n52n(n3),则当n1时,2a12a32a2n1()A BC D答案B解析由a5a2n52n(n3),得2an2n,ann.2a12a32a2n12232522n1.二、填空题5设f(x),利用
14、课本中推导等差数列前n项和的方法,可求得f(5)f(4)f(0)f(5)f(6)的值为_答案3解析f(0)f(1),f(x)f(1x),f(5)f(4)f(5)f(6)12(f(0)f(1)3.6求和1(13)(1332)(133232)(133n1)_.答案(3n1)解析a11,a213,a31332,an13323n1(3n1),原式(311)(321)(3n1)(3323n)n(3n1).三、解答题7(2013浙江理,18)在公差为d的等差数列an中,已知a110,且a1,2a22,5a3成等比数列(1)求d,an;(2)若d0,求|a1|a2|a3|an|.解析(1)由题意得a15a3
15、(2a22)2,a110,即d23d40.故d1或d4.所以ann11,nN*或an4n6,nN*.(2)设数列an的前n项和为Sn.因为d0,由(1)得d1,ann11.则当n11时,|a1|a2|a3|an|Snn2n.当n12时,|a1|a2|a3|an|Sn2S11n2n110.综上所述,|a1|a2|a3|an|8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的图象上,点(n,bn)在函数y2x的图象上(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)23(2n5)2n,2Tn322123(2n7)2n(2n5)2n1.两式相减得Tn6(23242n1)(2n5)2n1(2n5)2n16(72n)2n114.
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