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3√2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(q>0)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x≤0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d≠0,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(n∈N*,n≥2).(1)求证:数列{1an-3}是等差数列;(2)若a1=6,求数列{lgan}的前999项的和.11.(15分)设数列{an}满足a1=2,an+1-an=3·22n-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+5×42d=5×(-1)+5×4=15.故选A.4.D解析由3a1-a82+3a15=0,得a82=3a1+3a15=3(a1+a15)=3×2a8,即a82-6a8=0.因为a8=b10≠0,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),q>0,解得q=32,代入a1(1+q)=3a1q+2,解得a1=-1.6.C解析设x>0,则-x<0.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=a3×3+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t≥2,t∈N*,所以当t=3,即n=2时,(2Sn+16an+3)min=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n≥2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)]=4×n2=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,k∈N*,-2n+1,n=2k-1,k∈N*.10.(1)证明∵1an-3−1an-1-3=an-13an-1-9−1an-1-3=an-1-33an-1-9=13(n≥2),∴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n≥1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=n·22n-1知Sn=1·2+2·23+3·25+…+n·22n-1.①从而22·Sn=1·23+2·25+3·27+…+n·22n+1.②①-②,得(1-22)Sn=2+23+25+…+22n-1-n·22n+1,即Sn=19[(3n-1)22n+1+2].
