2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題突破練14 4.1-4.2 組合練 理

2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題突破練14 4.1-4.2 組合練 理一、選擇題(共9小題,滿分45分)1.我國古代數(shù)學(xué)名著算法統(tǒng)宗中有如下問題:“遠(yuǎn)望巍巍塔七層,紅光點(diǎn)點(diǎn)倍加增,共燈三百八十一,請問尖頭幾盞燈?”意思是:一座7層塔共掛了381盞燈,且相鄰兩層中的下一層燈數(shù)是上一層燈數(shù)的2倍,則塔的頂層共有燈() A.1盞B.3盞C.5盞D.9盞2.(2018遼寧大連二模,理4)設(shè)等比數(shù)列an的前n項(xiàng)和為Sn,S2=-1,S4=-5,則S6=()A.-9B.-21C.-25D.-633.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.604.(2018河北唐山三模,理6)數(shù)列an的首項(xiàng)a1=1,對于任意m,nN*,有an+m=an+3m,則an前5項(xiàng)和S5=()A.121B.25C.31D.355.(2018山東濰坊二模,理4)設(shè)數(shù)列an的前n項(xiàng)和為Sn,若Sn=-n2-n,則數(shù)列的前40項(xiàng)的和為()A.B.-C.D.-6.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.67.(2018吉林長春外國語學(xué)校二模,理8)已知數(shù)列an的前n項(xiàng)和Sn=2n-1,則數(shù)列的前10項(xiàng)和為()A.410-1B.(210-1)2C.(410-1)D.(210-1)8.設(shè)等差數(shù)列an滿足3a8=5a15,且a1>0,Sn為其前n項(xiàng)和,則數(shù)列Sn的最大項(xiàng)為()A.S23B.S24C.S25D.S269.(2018全國高考必刷模擬一,理11)數(shù)列an滿足a1=,an+1-1=an(an-1)(nN*),Sn=+,則Sn的整數(shù)部分的所有可能值構(gòu)成的集合是()A.0,1,2B.0,1,2,3C.1,2D.0,2二、填空題(共3小題,滿分15分)10.(2018湖南衡陽一模,文15)已知數(shù)列an的前n項(xiàng)和為Sn,若Sn=2an-2n,則Sn=. 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遼寧撫順一模,文16)已知數(shù)列an的前n項(xiàng)和為Sn,且a1=1,an+1=Sn+2,則a9的值為. 三、解答題(共3個題,分別滿分為13分,13分,14分)13.已知數(shù)列l(wèi)og2(an-1)(nN*)為等差數(shù)列,且a1=3,a3=9.(1)求數(shù)列an的通項(xiàng)公式;(2)證明:+<1.14.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)列的前n項(xiàng)和Tn.15.(2018河北保定一模,理17)已知數(shù)列an滿足:2an=an+1+an-1(n2,nN*),且a1=1,a2=2.(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足2anbn+1=an+1bn(n1,nN*),且b1=1.求數(shù)列bn的通項(xiàng)公式,并求其前n項(xiàng)和Tn.參考答案專題突破練144.14.2組合練1.B解析 設(shè)塔的頂層共有x盞燈,則各層的燈數(shù)構(gòu)成一個公比為2的等比數(shù)列,由=381,可得x=3,故選B.2.B解析 由題意,S2=a1+a2=-1,S4-S2=a3+a4=(a1+a2)q2=-4,q2=4,S6=S2+S4q2=-1+(-5)×4=-21.3.C解析 設(shè)等差數(shù)列an的公差為d,d0.由題意,得(a1+3d)2=(a1+2d)(a1+6d),即2a1+3d=0.S8=16,8a1+d=16,聯(lián)立解得a1=-,d=1.則S10=101=30.4.D解析 當(dāng)m=1時,由an+m=an+3m,得an+1-an=3,數(shù)列an是首項(xiàng)a1=1,公差d=3的等差數(shù)列,S5=5×1+5×4×3=35.5.D解析 Sn=-n2-n,a1=S1=-2.當(dāng)n2時,an=Sn-Sn-1=-n2-n+(n-1)2+(n-1)=-2n,a1=-2也滿足上式,則數(shù)列an的通項(xiàng)公式為an=-2n,=-,即數(shù)列的前40項(xiàng)的和為-+=-6.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+1=0,a1=-又=a1+m×1=3,-+m=3.m=5.故選C.7.C解析 Sn=2n-1,Sn+1=2n+1-1.an+1=Sn+1-Sn=(2n+1-1)-(2n-1)=2n.a1=S1=2-1=1,數(shù)列an的通項(xiàng)公式為an=2n-1,=4n-1,所求值為(410-1),故選C.8.C解析 設(shè)等差數(shù)列an的公差為d,3a8=5a15,3(a1+7d)=5(a1+14d),即2a1+49d=0.a1>0,d<0,等差數(shù)列an單調(diào)遞減.Sn=na1+d=nd=(n-25)2-d.當(dāng)n=25時,數(shù)列Sn取得最大值,故選C.9.A解析 a1=,an+1-1=an(an-1),an+1-an=(an-1)2>0,an+1>an,因此數(shù)列an單調(diào)遞增.an+1-1=an(an-1),Sn=+=3-由an+1-1=an(an-1)(nN*),得a2-1=,a2=,同理可得a3=,a4=當(dāng)n=1時,S1=3-,其整數(shù)部分為0,當(dāng)n=2時,S2=3-=3-=1+,其整數(shù)部分為1,當(dāng)n=2時,S3=3-=2+,其整數(shù)部分為2,因數(shù)列an單調(diào)遞增,當(dāng)n>4時,0<<1,所以當(dāng)n4時,Sn=3-(2,3),所以Sn的整數(shù)部分的所有可能值構(gòu)成的集合是0,1,2.10.n·2n解析 Sn=2an-2n=2(Sn-Sn-1)-2n,整理得Sn-2Sn-1=2n,等式兩邊同時除以2n,則=1.又S1=2a1-2=a1,可得a1=S1=2,數(shù)列是首項(xiàng)為1,公差為1的等差數(shù)列,所以=n,所以Sn=n·2n.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,解得a1q=8,q3=-,a8=a1q7=(a1q)(q3)2=8=2.12.384解析 當(dāng)n2時,由an+1=Sn+2,得an=Sn-1+2,兩式相減,得an+1-an=an,an+1=2an.當(dāng)n=2時,a2=S1+2=3,所以數(shù)列an中,當(dāng)n2時,是以2為公比的等比數(shù)列,a9=a2×27=3×128=384.13.(1)解 設(shè)等差數(shù)列l(wèi)og2(an-1)的公差為d.由a1=3,a3=9,得log22+2d=log28,即d=1.log2(an-1)=1+(n-1)×1=n,即an=2n+1.(2)證明 ,+=+=1-<1.14.解 (1)在3an=2Sn+3中,取n=1,得a1=3,且3an+1=2Sn+1+3,兩式相減,得3an+1-3an=2an+1,an+1=3an.a10,數(shù)列an是以3為公比的等比數(shù)列,an=3·3n-1=3n.(2)由(1)得bn=log3an=n,Tn=+=1-15.解 (1)由2an=an+1+an-1(n2,nN*),得數(shù)列an為等差數(shù)列,且首項(xiàng)為1,公差為a2-a1=1,所以an=n.(2)2nbn+1=(n+1)bn,(n1),數(shù)列是以=1為首項(xiàng),為公比的等比數(shù)列,即,從而bn=,Tn=+Tn=+,由-,得Tn=1+=2-,Tn=4-