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1、精編北師大版數(shù)學(xué)資料學(xué)業(yè)分層測(cè)評(píng)(八)(建議用時(shí):45分鐘)學(xué)業(yè)達(dá)標(biāo)一、選擇題1設(shè)an是公比為正數(shù)的等比數(shù)列,若a11,a516,則數(shù)列an前7項(xiàng)的和為()A63B64C127D128【解析】a5a1q4,q±2.q0,q2,S7127.【答案】C2等比數(shù)列an的前n項(xiàng)和為Sn,已知S3a210a1,a59,則a1()A.BC.D【解析】由S3a210a1,得a1a2a3a210a1,即a39a1,即a1q29a1,解得q29.又a59,a1q49,81a19,a1.【答案】C3在各項(xiàng)都為正數(shù)的等比數(shù)列an中,首項(xiàng)a13,a1a2a321,則a3a4a5()A33B72C84D189
2、【解析】a1a2a3a1a1qa1q2a1(1qq2)21,a13,1qq27,q3(舍)或q2,a3a4a5q2(a1a2a3)4×2184.【答案】C4(2016·吉安高二檢測(cè))在數(shù)列an中,已知對(duì)任意正整數(shù)n,有a1a2an2n1,則aaa等于()A(2n1)2B(2n1)2C4n1D(4n1)【解析】由a1a2an1an2n1,得a1a2an12n11,an2n1,a4n1,aaa(4n1)【答案】D5(2016·南昌高二檢測(cè))已知等比數(shù)列的公比為2,且前5項(xiàng)和為1,那么前10項(xiàng)和等于()A31B33C35D37【解析】法一:S51a1,S1033,故選B
3、.法二:a1a2a3a4a51,a6a7a8a9a10(a1a2a3a4a5)·q51×2532,S10a1a2a9a1013233.【答案】B二、填空題6若an是等比數(shù)列,且前n項(xiàng)和為Sn3n1t,則t_.【解析】法一:在等比數(shù)列an中,若q1,則Sn·qn,令A(yù),則SnAAqn.本題中Sn3n1t·3nt,t.法二:a1S1t1,a2S2S12,a3S3S26.an是等比數(shù)列,aa1·a3解得t.【答案】7等比數(shù)列an的前n項(xiàng)和為Sn,若S33S20,則公比q_.【解析】(1)顯然公比q1,設(shè)首項(xiàng)為a1,則由S33S20,得3×,
4、即q33q240,即q3q24q24q2(q1)4(q21)0,即(q1)(q24q4)0,所以q24q4(q2)20,解得q2.【答案】28已知等比數(shù)列an的前n項(xiàng)和為Sn,S540,S1080,則S15等于_【解析】因?yàn)镾5,S10S5,S15S10成等比數(shù)列,所以(S10S5)2S5·(S15S10),即(8040)240·(S1580),解得S15120.【答案】120三、解答題9在等比數(shù)列中,若Sn189,q2,an96,求a1和n.【解】由Sn及ana1·qn1得÷得,解得2n64,n6,代入得a13.10設(shè)an是等差數(shù)列,bn是各項(xiàng)都為正數(shù)
5、的等比數(shù)列,且a1b11,a3b521,a5b313.(1)求an,bn的通項(xiàng)公式(2)求數(shù)列的前n項(xiàng)和Sn.【解】(1)設(shè)an的公差為d,bn的公比為q,則依題意有q0,且解得所以an1(n1)d2n1,bnqn12n1.(2),Sn1,2Sn23,得Sn2222×22×6.能力提升1在等比數(shù)列an中,a14,q5,使Sn107的最小n值是()A11B10C12D9【解析】Sn5n1107,解得nlog5(1071)10log5(1071),nN,n11.【答案】A2已知等比數(shù)列an中,an2×3n1,則由此數(shù)列的偶數(shù)項(xiàng)所組成的新數(shù)列的前n項(xiàng)和Sn的值為() 【
6、導(dǎo)學(xué)號(hào):67940020】A3n1B3(3n1)C.D【解析】依據(jù)等比數(shù)列的性質(zhì)a2n也為等比數(shù)列,首項(xiàng)為6,公比為9,Sn(9n1)【答案】D3等比數(shù)列的前n項(xiàng)和Snm·3n2,則m_.【解析】設(shè)等比數(shù)列為an,則a1S13m2,S2a1a29m2,所以a26m,S3a1a2a327m2,所以a318m,又aa1·a3,所以(6m)2(3m2)·18m,解得m2或m0(舍),所以m2.【答案】24(2016·南昌高二檢測(cè))已知數(shù)列an是等比數(shù)列,且a1a2a36,a1·a2·a364,(|q|1)(1)求an的通項(xiàng)公式;(2)令bn
7、(2n1)·an,求數(shù)列bn的前n項(xiàng)和Sn.【解】(1)由a1a2a36,a1·a2·a364,得由于|q|1,解得a12,q2,所以an(2)n.(2)由(1)知bn(2n1)(2)n,Sn3·(2)5·(2)27·(2)3(2n1)·(2)n1(2n1)·(2)n,(2)·Sn3·(2)25·(2)37·(2)4(2n1)·(2)n(2n1)·(2)n1,得:3·Sn3·(2)2·(2)22·(2)32·(2)42·(2)n(2n1)·(2)n1,即3·Sn(2)2·(2)(2)2(2)3(2)4(2)n(2n1)·(2)n1,3·Sn(2)2·(2n1)·(2)n1,整理得:Sn·(2)n1.