2019版高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 專題對點(diǎn)練15 4.1~4.2組合練 文.doc

專題對點(diǎn)練154.14.2組合練(限時90分鐘,滿分100分)一、選擇題(共9小題,滿分45分)1.設(shè)Sn是等差數(shù)列an的前n項(xiàng)和,若a1+a3+a5=3,則S5=()A.5B.7C.9D.112.九章算術(shù)是我國第一部數(shù)學(xué)專著,下有源自其中的一個問題:“今有金箠,長五尺,斬本一尺,重四斤,斬末一尺,重二斤.問金箠重幾何?”其意思為:今有金杖(粗細(xì)均勻變化)長5尺,截得本端1尺,重4斤,截得末端1尺,重2斤,則金杖重()A.18斤B.15斤C.13斤D.20斤3.已知等差數(shù)列an的公差為2,若a2,a4,a8成等比數(shù)列,則an的前n項(xiàng)和Sn=()A.n(n+1)B.n(n-1)C.n(n+1)2D.n(n-1)24.公差不為零的等差數(shù)列an的前n項(xiàng)和為Sn.若a4是a3與a7的等比中項(xiàng),S8=16,則S10等于()A.18B.24C.30D.605.等比數(shù)列an的前n項(xiàng)和為Sn.已知S3=a2+10a1,a5=9,則a1=()A.B.-C.D.-6.(2018廣東深圳耀華模擬)在數(shù)列an中,a1=1,an+1=2an-2n,則a17=()A.-15216B.15217C.-16216D.162177.設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,則m=()A.3B.4C.5D. 68.在等比數(shù)列an中,各項(xiàng)均為正數(shù),Sn是其前n項(xiàng)和,且滿足2S3=8a1+3a2,a4=16,則S4=()A.9B.15C.18D.309.在遞減等差數(shù)列an中, a1a3=a22-4.若a1=13,則數(shù)列1anan+1的前n項(xiàng)和的最大值為()A.24143B.1143C.2413D.613二、填空題(共3小題,滿分15分)10.已知等比數(shù)列an,a2a4=a5,a4=8,則an的前4項(xiàng)和S4=.11.設(shè)等比數(shù)列an的前n項(xiàng)和為Sn,若S3,S9,S6成等差數(shù)列,且a2+a5=4,則a8的值為.12.(2018湖北重點(diǎn)高中協(xié)作體模擬)定義“等積數(shù)列”,在一個數(shù)列中,如果每一項(xiàng)與它的后一項(xiàng)的積都為同一個常數(shù),那么這個數(shù)列叫做等積數(shù)列,這個常數(shù)叫做該數(shù)列的公積.已知數(shù)列an是等積數(shù)列且a1=2,公積為10,則這個數(shù)列前21項(xiàng)和S21的值為.三、解答題(共3個題,滿分分別為13分,13分,14分)13.已知數(shù)列an的前n項(xiàng)和為Sn,且對任意正整數(shù)n,都有3an=2Sn+3成立.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=log3an,求數(shù)列bn的前n項(xiàng)和Tn.14.已知數(shù)列an的前n項(xiàng)和為Sn,且滿足Sn+n=2an(nN*).(1)證明:數(shù)列an+1為等比數(shù)列,并求數(shù)列an的通項(xiàng)公式;(2)若bn=(2n+1)an+2n+1,數(shù)列bn的前n項(xiàng)和為Tn,求滿足不等式Tn-22n-1>2 010的n的最小值.15.已知數(shù)列an的前n項(xiàng)和為Sn,且滿足a1=1,2Sn=(n+1)an.在數(shù)列bn中,bn=2an+1.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)求數(shù)列1anlog2bn的前n項(xiàng)和Tn.專題對點(diǎn)練15答案1.A解析 由a1+a3+a5=3,得3a3=3,解得a3=1.故S5=5(a1+a5)2=5a3=5.2.B解析 由題意可知,在等差數(shù)列an中,a1=4,a5=2,則S5=(a1+a5)52=(4+2)52=15,故金杖重15斤.3.A解析 a2,a4,a8成等比數(shù)列,a42=a2a8,即(a1+6)2=(a1+2)(a1+14),解得a1=2.Sn=na1+n(n-1)2d=2n+n2-n=n2+n=n(n+1).故選A.4.C解析 設(shè)等差數(shù)列an的公差為d0.由題意,得(a1+3d)2=(a1+2d)(a1+6d),即2a1+3d=0.S8=16,8a1+872d=16,聯(lián)立解得a1=-,d=1.則S10=10-32+10921=30.5.C解析 設(shè)數(shù)列an的公比為q,若q=1,則由a5=9,得a1=9,此時S3=27,而a2+10a1=99,不滿足題意,因此q1.當(dāng)q1時,S3=a1(1-q3)1-q=a1q+10a1,1-q31-q=q+10,整理得q2=9.a5=a1q4=9,即81a1=9,a1=.6.A解析 由題意可得an+12n+1=an2n-12,即an+12n+1-an2n=-,據(jù)此可得,數(shù)列an2n是首項(xiàng)為a121=12,公差為-的等差數(shù)列,故a17217=12+(17-1)-12=-152,a17=-15216.故選A.7.C解析 Sm-1=-2,Sm=0,Sm+1=3,am=Sm-Sm-1=0-(-2)=2,am+1=Sm+1-Sm=3-0=3.d=am+1-am=3-2=1.Sm=ma1+m(m-1)21=0,a1=-m-12.又am+1=a1+m1=3,-m-12+m=3.m=5.故選C.8.D解析 設(shè)等比數(shù)列an的公比為q>0,2S3=8a1+3a2,2(a1+a2+a3)=8a1+3a2,即2a1q2=6a1+a1q,即2q2-q-6=0,解得q=2.又a4=16,可得a123=16,解得a1=2.則S4=2(1-24)1-2=30.9.D解析 設(shè)公差為d,則d<0.由題意,得13(13+2d)=(13+d)2-4,解得d=-2或d=2(舍去),an=a1+(n-1)d=15-2n.當(dāng)an=15-2n0時,即n7.5;當(dāng)an+1=13-2n0時,即n6.5.當(dāng)n7時,an>0.1anan+1=1(15-2n)(13-2n)=1212n-15-12n-13,數(shù)列1anan+1的前n項(xiàng)和為121-13-1-11+1-11-1-9+12n-15-12n-13=12-113-12n-13,當(dāng)n=6時,數(shù)列1anan+1的前n項(xiàng)和最大,最大值為12-113+1=613,故選D.10.15解析 設(shè)等比數(shù)列an的公比為q,a2a4=a1qa4=a1a5=a5,a1=1.又a4=8,q3=8,q=2.故S4=1(1-24)1-2=15.11.2解析 等比數(shù)列an的前n項(xiàng)和為Sn,S3,S9,S6成等差數(shù)列,且a2+a5=4,2a1(1-q9)1-q=a1(1-q3)1-q+a1(1-q6)1-q,a1q+a1q4=4,解得a1q=8,q3=-,a8=a1q7=(a1q)(q3)2=8=2.12.72解析 由數(shù)列an是等積數(shù)列,且 a1=2,公積為10,根據(jù)等積數(shù)列的定義,得a2=5,a3=2,由此可以知道數(shù)列an的所有奇數(shù)項(xiàng)為2,所有偶數(shù)項(xiàng)為5.故這個數(shù)列前21項(xiàng)和S21=710+2=72.13.解 (1)在3an=2Sn+3中,令n=1,得a1=3.當(dāng)n2時,3an=2Sn+3,3an-1=2Sn-1+3,-得an=3an-1,數(shù)列an是以3為首項(xiàng),3為公比的等比數(shù)列,an=3n.(2)由(1)得bn=log3an=n,數(shù)列bn的前n項(xiàng)和Tn=1+2+3+n=n(n+1)2.14.(1)證明 當(dāng)n=1時,2a1=a1+1,a1=1.2an=Sn+n,nN*,2an-1=Sn-1+n-1,n2,兩式相減,得an=2an-1+1,n2,即an+1=2(an-1+1),n2,數(shù)列an+1為以2為首項(xiàng),2為公比的等比數(shù)列,an+1=2n,an=2n-1,nN*.(2)解 bn=(2n+1)an+2n+1=(2n+1)2n,Tn=32+522+(2n+1)2n,2Tn=322+523+(2n+1)2n+1,兩式相減可得-Tn=32+222+223+22n-(2n+1)2n+1,Tn=(2n-1)2n+1+2,Tn-22n-1>2 010 可化為2n+1>2 010.210=1 024,211=2 048,滿足不等式Tn-22n-1>2 010的n的最小值為10.15.解 (1)當(dāng)n2時,由2Sn=(n+1)an,得2Sn-1=nan-1,兩式相減得2an=(n+1)an-nan-1,整理得anan-1=nn-1.由an=anan-1an-1an-2a2a1=nn-1n-1n-21=n(n2).又當(dāng)n=1時,a1=1,an=n(nN*).由bn=2an+1=2n+1,bn的通項(xiàng)公式為bn=2n+1.(2)由(1)得1anlog2bn=1nlog22n+1=1n(n+1)=1n-1n+1.Tn=1-12+12-13+1n-1n+1=1-12+12-13+1n-1n+1=1-1n+1=nn+1.故數(shù)列1anlog2bn的前n項(xiàng)和Tn=nn+1.