2020版高考數(shù)學(xué)一輪復(fù)習(xí) 第六章 數(shù)列單元質(zhì)檢卷6B 文 北師大版.doc

單元質(zhì)檢卷六數(shù)列(B)(時(shí)間:45分鐘滿分:100分)一、選擇題(本大題共6小題,每小題7分,共42分)1.(2018廣東汕頭潮南5月沖刺)等比數(shù)列an的前n項(xiàng)和為Sn,4a1,2a2,a3成等差數(shù)列,a1=1,則S4=()A.15B.-15C.4D.-42.(2018山東濰坊青州三模,7)已知等差數(shù)列an的前n項(xiàng)和為Sn,若a3+a4+a11=18,則S11=()A.9B.22C.36D.663.已知an為等差數(shù)列,bn為正項(xiàng)等比數(shù)列,公比q1,若a1=b1,a9=b9,則()A.a5=b5B.a5>b5C.a5<b5D.以上都有可能4.(2018遼寧大連二模,6)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.-635.(2018江西教學(xué)質(zhì)量監(jiān)測(cè),9)已知等比數(shù)列an的首項(xiàng)a1=2,前n項(xiàng)和為Sn,若S5+4S3=5S4,則數(shù)列2log2an+1log2an-6的最大項(xiàng)等于()A.-11B.-C.193D.156.(2018河北衡水八模,8)已知函數(shù)f(x)=ax+b(a>0,a1)的圖像經(jīng)過(guò)點(diǎn)P(1,3),Q(2,5).當(dāng)nN+時(shí),an=f(n)-1f(n)f(n+1),記數(shù)列an的前n項(xiàng)和為Sn,當(dāng)Sn=1033時(shí),n的值為()A.7B.6C.5D.4二、填空題(本大題共2小題,每小題7分,共14分)7.(2018吉林實(shí)驗(yàn)中學(xué)期末)已知在公比q>1的等比數(shù)列an中,a2+a3=12,a1a4=32,數(shù)列bn滿足bn=log2an,則數(shù)列bn的前10項(xiàng)和S10=.8.(2018河南六市聯(lián)考一,16)已知正項(xiàng)數(shù)列an的前n項(xiàng)和為Sn,若an和Sn都是等差數(shù)列,且公差相等,則a2=.三、解答題(本大題共3小題,共44分)9.(14分)(2018北京西城一模,15)設(shè)等差數(shù)列an的公差不為0,a2=1,且a2,a3,a6成等比數(shù)列.(1)求an的通項(xiàng)公式;(2)設(shè)數(shù)列an的前n項(xiàng)和為Sn,求使Sn>35成立的n的最小值.10.(15分)(2018山東師大附中一模,17)已知等差數(shù)列an是遞增數(shù)列,且滿足a4a7=15,a3+a8=8.(1)求數(shù)列an的通項(xiàng)公式;(2)令bn=19an-1an(n2),b1=,求數(shù)列bn的前n項(xiàng)和Sn.11.(15分)(2018寧夏銀川一中一模,17)設(shè)Sn為數(shù)列an的前n項(xiàng)和,已知an>0,an2+2an=4Sn+3.(1)求an的通項(xiàng)公式;(2)設(shè)bn=1anan+1,求數(shù)列bn的前n項(xiàng)和.單元質(zhì)檢卷六數(shù)列(B)1.A4a1,2a2,a3成等差數(shù)列且a1=1,4a1+a3=22a2,即4+q2-4q=0,解得q=2,a2=2,a3=4,a4=8,S4=1+2+4+8=15.故選A.2.Da3+a4+a11=18,3a1+15d=18a1+5d=6,S11=11(a1+5d)=116=66,故選D.3.B由等差、等比中項(xiàng)的定義可知a5=a1+a92,b5=b1b9.又a1=b1,a9=b9,所以a1+a92a1a9=b1b9,即a5>b5,故選B.4.B由題意,S2=a1+a2=-1,S4-S2=a3+a4=(a1+a2)q2=-4,q2=4,S6=S2+S4q2=-1+(-5)4=-21.5.D由已知得,q1.由S5+4S3=5S4得1-q51-q+41-q31-q=51-q41-q,解得q=4.an=24n-1=22n-1,2log2an+1log2an-6=4n-12n-7,由函數(shù)y=4x-12x-7=2+132x-7的圖像得到,當(dāng)n=4時(shí),數(shù)列2log2an+1log2an-6的最大項(xiàng)等于15.6.D由題意,得a+b=3,a2+b=5,a=2,b=1,f(x)=2x+1.an=2n+1-1(2n+1)(2n+1+1)=12n+1-12n+1+1.Sn=13-15+15-17+12n+1-12n+1+1=13-12n+1+1=1033,2n+1+1=33,n=4,故選D.7.55因?yàn)樵诘缺葦?shù)列an中,a2+a3=12,a1a4=32,所以a2+a3=12,a2a3=32,解得a2=4,a3=8,或a2=8,a3=4.又q>1,所以a2=4,a3=8,所以q=2.所以an=a2qn-2=2n,bn=log22n=n,則S10=10(1+10)2=55.8.設(shè)數(shù)列an的公差為d(d>0),又Sn也是公差為d的等差數(shù)列,則S2=2a1+d=a1+d,兩邊平方得2a1+d=a1+2da1+d2,S3=3a1+3d=a1+2d,兩邊平方得3a1+3d=a1+4da1+4d2,-得a1=-2d+2da1+3d2,把代入得d(2d-1)=0.所以d=0或d=12.當(dāng)d=0時(shí),a1=0,不合題意,當(dāng)d=12時(shí),代入解得a1=14.所以a2=a1+d=34.9.解 (1)設(shè)等差數(shù)列an的公差為d,d0.a2,a3,a6成等比數(shù)列,a32=a2a6,即(1+d)2=1+4d,解得d=2,或d=0(舍去),an=a2+(n-2)d=2n-3.(2)an=2n-3,Sn=n(a1+an)2=n(a2+an-1)2=n2-2n.依題意有n2-2n>35,解得n>7.因此使Sn>35成立的n的最小值為8.10.解 (1)a4a7=15,a3+a8=a4+a7=8,解得a4=3,a7=5,d=23,an=1+23(n-1)=23n+13.(2)bn=19an-1an=1(2n-1)(2n+1)=12(12n-1-12n+1)(n2),b1=13=121-13滿足上式,bn的通項(xiàng)公式為bn=1212n-1-12n+1.Sn=121-13+13-15+12n-1-12n+1=121-12n+1=n2n+1.11.解 (1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.兩式相減,得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).an>0,an+1-an=2,a12+2a1=4a1+3,a1=-1(舍)或a1=3,則an是首項(xiàng)為3,公差d=2的等差數(shù)列,an的通項(xiàng)公式an=3+2(n-1)=2n+1.(2)an=2n+1,bn=1anan+1=1(2n+1)(2n+3)=1212n+1-12n+3,數(shù)列bn的前n項(xiàng)和Tn=1213-15+15-17+12n+1-12n+3=1213-12n+3=n3(2n+3).