廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 單元質(zhì)檢六 數(shù)列（A） 文.docx

單元質(zhì)檢六數(shù)列(A)(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.已知等差數(shù)列an的前n項(xiàng)和為Sn,a6=15,S9=99,則等差數(shù)列an的公差是()A.14B.4C.-4D.-3答案B解析數(shù)列an是等差數(shù)列,a6=15,S9=99,a1+a9=22,2a5=22,a5=11.公差d=a6-a5=4.2.已知公比為32的等比數(shù)列an的各項(xiàng)都是正數(shù),且a3a11=16,則log2a16=()A.4B.5C.6D.7答案B解析由等比中項(xiàng)的性質(zhì),得a3a11=a72=16.因?yàn)閿?shù)列an各項(xiàng)都是正數(shù),所以a7=4.所以a16=a7q9=32.所以log2a16=5.3.在等差數(shù)列an中,已知a4=5,a3是a2和a6的等比中項(xiàng),則數(shù)列an的前5項(xiàng)的和為()A.15B.20C.25D.15或25答案A解析設(shè)an的公差為d.在等差數(shù)列an中,a4=5,a3是a2和a6的等比中項(xiàng),a1+3d=5,(a1+2d)2=(a1+d)(a1+5d),解得a1=-1,d=2,S5=5a1+542d=5(-1)+54=15.故選A.4.已知等差數(shù)列an和等比數(shù)列bn滿足3a1-a82+3a15=0,且a8=b10,則b3b17=()A.9B.12C.16D.36答案D解析由3a1-a82+3a15=0,得a82=3a1+3a15=3(a1+a15)=32a8,即a82-6a8=0.因?yàn)閍8=b100,所以a8=6,b10=6,所以b3b17=b102=36.5.設(shè)公比為q(q>0)的等比數(shù)列an的前n項(xiàng)和為Sn,若S2=3a2+2,S4=3a4+2,則a1=()A.-2B.-1C.12D.23答案B解析S2=3a2+2,S4=3a4+2,S4-S2=3(a4-a2),即a1(q3+q2)=3a1(q3-q),q>0,解得q=32,代入a1(1+q)=3a1q+2,解得a1=-1.6.已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x0時(shí),f(x)=x(1-x).若數(shù)列an滿足a1=12,且an+1=11-an,則f(a11)=()A.2B.-2C.6D.-6答案C解析設(shè)x>0,則-x<0.因?yàn)閒(x)是定義在R上的奇函數(shù),所以f(x)=-f(-x)=-x(1+x)=x(1+x).由a1=12,且an+1=11-an,得a2=11-a1=11-12=2,a3=11-a2=11-2=-1,a4=11-a3=11-(-1)=12,所以數(shù)列an是以3為周期的周期數(shù)列,即a11=a33+2=a2=2.所以f(a11)=f(a2)=f(2)=2(1+2)=6.二、填空題(本大題共2小題,每小題7分,共14分)7.(2018福建莆田質(zhì)檢)已知數(shù)列an滿足a1=1,an-an+1=2anan+1,則a6=.答案111解析由an-an+1=2anan+1,得1an+1-1an=2,即數(shù)列1an是以1a1=1為首項(xiàng),2為公差的等差數(shù)列.所以1a6=1a1+52=11,即a6=111.8.已知等比數(shù)列an滿足a2+8a5=0,設(shè)Sn是數(shù)列1an的前n項(xiàng)和,則S5S2=.答案-11解析設(shè)an的公比為q.由a2+8a5=0,得a1q+8a1q4=0,解得q=-12.易知1an是等比數(shù)列,公比為-2,首項(xiàng)為1a1,所以S2=1a11-(-2)21-(-2)=-1a1,S5=1a11-(-2)51-(-2)=11a1,所以S5S2=-11.三、解答題(本大題共3小題,共44分)9.(14分)記Sn為等差數(shù)列an的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求an的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解(1)設(shè)an的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.所以an的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.10.(15分)已知數(shù)列an滿足an=6-9an-1(nN*,n2).(1)求證:數(shù)列1an-3是等差數(shù)列;(2)若a1=6,求數(shù)列l(wèi)g an的前999項(xiàng)的和.(1)證明1an-3-1an-1-3=an-13an-1-9-1an-1-3=an-1-33an-1-9=13(n2),數(shù)列1an-3是等差數(shù)列.(2)解1an-3是等差數(shù)列,且1a1-3=13,d=13,1an-3=1a1-3+13(n-1)=n3.an=3(n+1)n.lgan=lg(n+1)-lgn+lg3.設(shè)數(shù)列l(wèi)gan的前999項(xiàng)的和為S,則S=999lg3+(lg2-lg1+lg3-lg2+lg1000-lg999)=999lg3+lg1000=3+999lg3.11.(15分)設(shè)數(shù)列an滿足a1=2,an+1-an=322n-1.(1)求數(shù)列an的通項(xiàng)公式;(2)令bn=nan,求數(shù)列bn的前n項(xiàng)和Sn.解(1)由已知,當(dāng)n1時(shí),an+1=(an+1-an)+(an-an-1)+(a2-a1)+a1=3(22n-1+22n-3+2)+2=22(n+1)-1.而a1=2,所以數(shù)列an的通項(xiàng)公式為an=22n-1.(2)由bn=nan=n22n-1知Sn=12+223+325+n22n-1.從而22Sn=123+225+327+n22n+1.-,得(1-22)Sn=2+23+25+22n-1-n22n+1,即Sn=19(3n-1)22n+1+2.