单元质检五数列(A)(时间:45分钟满分:100分)一、选择题(本大题共6小题,每小题7分,共42分)1.已知等差数列an的前n项和为Sn,a6=15,S9=99,则等差数列an的公差是()A.14B.4C.-4D.-32.已知公比为32的等比数列an的各项都是正数,且a3a11=16,则log2a16=()A.4B.5C.6D.73.在等差数列an中,已知a4=5,a3是a2和a6的等比中项,则数列an的前5项的和为()A.15B.20C.25D.15或254.已知等差数列an和等比数列bn满足:3a1-a82+3a15=0,且a8=b10,则b3b17=()A.9B.12C.16D.365.设公比为q(q0)的等比数列an的前n项和为Sn,若S2=3a2+2,S4=3a4+2,则a1=()A.-2B.-1C.12D.236.已知函数f(x)是定义在R上的奇函数,当x0时,f(x)=x(1-x).若数列an满足a1=12,且an+1=11-an,则f(a11)=()A.2B.-2C.6D.-6二、填空题(本大题共2小题,每小题7分,共14分)7.已知Sn为等比数列an的前n项和,且Sn=2an-1,则数列an的公比q=.8.已知等差数列an的公差d0,且a1,a3,a13成等比数列,若a1=1,Sn为数列an的前n项和,则2Sn+16an+3的最小值为.三、解答题(本大题共3小题,共44分)9.(14分)已知数列an的首项为a1=1,其前n项和为Sn,且数列Snn是公差为2的等差数列.(1)求数列an的通项公式;(2)若bn=(-1)nan,求数列bn的前n项和Tn.10.(15分)已知数列an满足an=6-9an-1(nN*,n2).(1)求证:数列1an-3是等差数列;(2)若a1=6,求数列lg an的前999项的和.11.(15分)设数列an满足a1=2,an+1-an=322n-1.(1)求数列an的通项公式;(2)令bn=nan,求数列bn的前n项和Sn.单元质检五数列(A)1.B解析数列an是等差数列,a6=15,S9=99,a1+a9=22,2a5=22,a5=11.公差d=a6-a5=4.2.B解析由等比中项的性质,得a3a11=a72=16.因为数列an各项都是正数,所以a7=4.所以a16=a7q9=32.所以log2a16=5.3.A解析在等差数列an中,a4=5,a3是a2和a6的等比中项,a1+3d=5,(a1+2d)2=(a1+d)(a1+5d),解得a1=-1,d=2,S5=5a1+542d=5(-1)+54=15.故选A.4.D解析由3a1-a82+3a15=0,得a82=3a1+3a15=3(a1+a15)=32a8,即a82-6a8=0.因为a8=b100,所以a8=6,b10=6,所以b3b17=b102=36.5.B解析S2=3a2+2,S4=3a4+2,S4-S2=3(a4-a2),即a1(q3+q2)=3a1(q3-q),q0,解得q=32,代入a1(1+q)=3a1q+2,解得a1=-1.6.C解析设x0,则-x0.因为f(x)是定义在R上的奇函数,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,所以数列an是以3为周期的周期数列,即a11=a33+2=a2=2.所以f(a11)=f(a2)=f(2)=2(1+2)=6.7.2解析Sn=2an-1,a1=2a1-1,a1+a2=2a2-1,解得a1=1,a2=2.等比数列an的公比q=2.8.4解析设an的公差为d.因为a1,a3,a13成等比数列,所以(1+2d)2=1+12d,解得d=2.所以an=2n-1,Sn=n2.所以2Sn+16an+3=2n2+162n+2=n2+8n+1.令t=n+1,则原式=t2+9-2tt=t+9t-2.因为t2,tN*,所以当t=3,即n=2时,2Sn+16an+3min=4.9.解(1)数列Snn是公差为2的等差数列,且S11=a1=1,Snn=1+(n-1)2=2n-1.Sn=2n2-n.当n2时,an=Sn-Sn-1=2n2-n-2(n-1)2-(n-1)=4n-3.a1符合an=4n-3,an=4n-3.(2)由(1)可得bn=(-1)nan=(-1)n(4n-3).当n为偶数时,Tn=(-1+5)+(-9+13)+-(4n-7)+(4n-3)=4n2=2n;当n为奇数时,n+1为偶数,Tn=Tn+1-bn+1=2(n+1)-(4n+1)=-2n+1.综上所述,Tn=2n,n=2k,kN*,-2n+1,n=2k-1,kN*.10.(1)证明1an-31an-1-3=an-13an-1-91an-1-3=an-1-33an-1-9=13(n2),数列1an-3是等差数列.(2)解1an-3是等差数列,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.设数列lgan的前999项的和为S,则S=999lg3+(lg2-lg1+lg3-lg2+lg1000-lg999)=999lg3+lg1000=3+999lg3.11.解(1)由已知,当n1时,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,所以数列an的通项公式为an=22n-1.(2)由bn=nan=n22n-1知Sn=12+223+325+n22n-1.从而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.