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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)(九)(建議用時(shí):45分鐘)學(xué)業(yè)達(dá)標(biāo)一、選擇題1數(shù)列12n1的前n項(xiàng)和為()A12nB22nCn2n1Dn22n【解析】Snnn2n1.【答案】C2若數(shù)列an的通項(xiàng)公式是an(1)n(3n2),則a1a2a10()A15B12 C12D15【解析】設(shè)bn3n2,則數(shù)列bn是以1為首項(xiàng),3為公差的等差數(shù)列,所以a1a2a9a10(b1)b2(b9)b10(b2b1)(b4b3)(b10b9)5×315.【答案】A3數(shù)列1,3,5,7,的前n項(xiàng)和Sn為()An21Bn22Cn21Dn22【解析】由題意知數(shù)列的通項(xiàng)為an2n1,則Snn21.【答案】C4已知
2、數(shù)列an的通項(xiàng)公式是an,若前n項(xiàng)和為10,則項(xiàng)數(shù)n為()A11B99C120D121【解析】an,Sna1a2an(1)()()1.令110,得n120.【答案】C5數(shù)列1,的前n項(xiàng)和為()A.BC.D【解析】該數(shù)列的通項(xiàng)為an,分裂為兩項(xiàng)差的形式為an2,則Sn2,Sn2.【答案】B二、填空題6某住宅小區(qū)計(jì)劃植樹(shù)不少于100棵,若第一天植2棵,以后每天植樹(shù)的棵樹(shù)是前一天的2倍,則需要的最少天數(shù)n(nN*)的值為_(kāi)【解析】由題意可得,第n天種樹(shù)的棵數(shù)an是以2為首項(xiàng),以2為公比的等比數(shù)列,Sn2n12100,2n1102.nN*,n17,n6,即n的最小值為6.【答案】67已知數(shù)列an的前n
3、項(xiàng)和為Sn且ann·2n,則Sn_.【解析】ann·2n,Sna1a2an1·22·223·23n·2n,2Sn1·222·233·24(n1)·2nn·2n1,得Sn222232nn·2n1n·2n12n12n·2n1.Sn(n1)·2n12.【答案】(n1)·2n128已知等比數(shù)列an中,a13,a481,若數(shù)列bn滿足bnlog3an,則數(shù)列的前n項(xiàng)和Sn_.【解析】設(shè)等比數(shù)列an的公比為q,則q327,解得q3,所以ana1qn
4、13×3n13n,故bnlog3ann,所以,則Sn11.【答案】三、解答題9已知等差數(shù)列an的前3項(xiàng)和為6,前8項(xiàng)和為4.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn(4an)qn1(q0,nN*),求數(shù)列bn的前n項(xiàng)和Sn. 【導(dǎo)學(xué)號(hào):67940023】【解】(1)設(shè)an的公差為d,則由已知得即解得a13,d1,故an3(n1)4n(nN*)(2)由(1)知,bnn·qn1,于是Sn1·q02·q13·q2n·qn1,若q1,上式兩邊同乘以q,qSn1·q12·q2(n1)·qn1n·qn,兩式
5、相減得(1q)Sn1q1q2qn1n·qnn·qn.Sn.若q1,則Sn123n,Sn10等差數(shù)列an的各項(xiàng)均為正數(shù),a13,前n項(xiàng)和為Sn,bn為等比數(shù)列,b11,且b2S264,b3S3960.(1)求an與bn;(2)求.【解】(1)設(shè)an的公差為d,bn的公比為q,則d為正數(shù),an3(n1)d,bnqn1.依題意有解得或(舍去)故an32(n1)2n1,bn8n1.(2)Sn35(2n1)n(n2),所以.能力提升1(2016·金華高二檢測(cè))已知an為等比數(shù)列,Sn是它的前n項(xiàng)和若a2·a32a1,且a4與2a7的等差中項(xiàng)為,則S5()A35B3
6、3C31D29【解析】設(shè)an的公比為q,則由等比數(shù)列的性質(zhì)知,a2·a3a1·a42a1,即a42.由a4與2a7的等差中項(xiàng)為知,a42a72×,a7,q3,即q,a4a1q3a1×2,a116,S531.【答案】C2已知數(shù)列an滿足an1,且a1,則該數(shù)列的前2 016項(xiàng)的和等于()A1 509B3 018 C1 512D2 016【解析】因?yàn)閍1,又an1,所以a21,從而a3,a41,即得an故數(shù)列的前2 016項(xiàng)的和等于S2 0161 008×1 512.【答案】C3已知數(shù)列an:,那么數(shù)列bn的前n項(xiàng)和Sn為_(kāi)【解析】由已知條件可得數(shù)列an的通項(xiàng)為an,bn4.Sn44.【答案】4(2014·湖南高考)已知數(shù)列an的前n項(xiàng)和Sn,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn2an(1)nan,求數(shù)列bn的前2n項(xiàng)和【解】(1)當(dāng)n1時(shí),a1S11;當(dāng)n2時(shí),anSnSn1n.故數(shù)列an的通項(xiàng)公式為ann.(2)由(1)知ann,故bn2n(1)nn.記數(shù)列bn的前2n項(xiàng)和為T2n,則T2n(212222n)(12342n)記A212222n,B12342n,則A22n12,B(12)(34)(2n1)2nn,故數(shù)列bn的前2n項(xiàng)和T2nAB22n1n2.