（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 單元質(zhì)檢5 數(shù)列（B）（含解析）新人教A版

單元質(zhì)檢五　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.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a3=5,S6=36,則a6=(　　)A.9 B.10 C.11 D.122.在單調(diào)遞減的等比數(shù)列{an}中,若a3=1,a2+a4=52,則a1=(　　)A.2 B.4 C.2 D.223.設(shè)an=-n2+9n+10,則數(shù)列{an}前n項(xiàng)和最大時(shí)n的值為(　　)A.9 B.10C.9或10 D.124.等比數(shù)列{an}的前n項(xiàng)和為Sn,已知S3=a2+5a1,a7=2,則a5=(　　)A.12 B.-12 C.2 D.-25.對(duì)于數(shù)列{an},定義數(shù)列{an+1-an}為數(shù)列{an}的“差數(shù)列”,若a1=2,數(shù)列{an}的“差數(shù)列”的通項(xiàng)為2n,則數(shù)列{an}的前n項(xiàng)和Sn=(　　)A.2 B.2nC.2n+1-2 D.2n-1-26.已知函數(shù)f(x)=ax+b(a>0,a≠1)的圖象經(jīng)過(guò)點(diǎn)P(1,3),Q(2,5).當(dāng)n∈N*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.7 B.6 C.5 D.4二、填空題(本大題共2小題,每小題7分,共14分)7.在3和一個(gè)未知數(shù)之間填上一個(gè)數(shù),使三個(gè)數(shù)成等差數(shù)列,若中間項(xiàng)減去6,則三個(gè)數(shù)成等比數(shù)列,此未知數(shù)是　　　　　　　　.?8.已知數(shù)列{an}滿足:a1=1,an=an-12+2an-1(n≥2),若bn=1an+1+1an+2(n∈N*),則數(shù)列{bn}的前n項(xiàng)和Sn=　　　　　.?三、解答題(本大題共3小題,共44分)9.(14分)已知各項(xiàng)均為正數(shù)的等比數(shù)列{an},其前n項(xiàng)和為Sn,S2=6,S4=30.(1)求{an}的通項(xiàng)公式;(2)設(shè)bn=1(log2an)(log2an+2),{bn}的前n項(xiàng)和為Tn,證明:Tn<34.10.(15分)已知{an}為等差數(shù)列,前n項(xiàng)和為Sn(n∈N*),{bn}是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.(1)求{an}和{bn}的通項(xiàng)公式;(2)求數(shù)列{a2nb2n-1}的前n項(xiàng)和(n∈N*).11.(15分)已知數(shù)列{an}滿足a1=1,an+1=1-14an,其中n∈N*.(1)設(shè)bn=22an-1,求證:數(shù)列{bn}是等差數(shù)列,并求出{an}的通項(xiàng)公式.(2)設(shè)cn=4ann+1,數(shù)列{cncn+2}的前n項(xiàng)和為Tn,是否存在正整數(shù)m,使得Tn<1cmcm+1對(duì)于n∈N*恒成立?若存在,求出m的最小值;若不存在,請(qǐng)說(shuō)明理由.單元質(zhì)檢五　數(shù)列(B)1.C　解析∵S6=a1+a62×6=(a3+a4)2×6=36,又a3=5,∴a4=7.∴a6=a4+(6-4)×(7-5)=11.故選C.2.B　解析由已知,得a1q2=1,a1q+a1q3=52,∴q+q3q2=52,q2-52q+1=0,∴q=12(q=2舍去),∴a1=4.3.C　解析令an≥0,得n2-9n-10≤0,∴1≤n≤10.令an+1≤0,即n2-7n-18≥0,∴n≥9.∴9≤n≤10.∴前9項(xiàng)和等于前10項(xiàng)和,它們都最大.4.A　解析由條件,得a1+a2+a3=a2+5a1,a7=2,∴a1q2=4a1,a1q6=2,∴q2=4,a1=132,∴a5=a1q4=132×42=12.5.C　解析∵an+1-an=2n,∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2n-1+2n-2+…+22+2+2=2-2n1-2+2=2n-2+2=2n.∴Sn=2-2n+11-2=2n+1-2.故選C.6.D　解析∵函數(shù)f(x)=ax+b(a>0,a≠1)的圖象經(jīng)過(guò)點(diǎn)P(1,3),Q(2,5),∴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,∴n=4,故選D.7.3或27　解析設(shè)此三個(gè)數(shù)為3,a,b,則2a=3+b,(a-6)2=3b,解得a=3,b=3或a=15,b=27.故這個(gè)未知數(shù)為3或27.8.1-122n-1　解析當(dāng)n≥2時(shí),an+1=an-12+2an-1+1=(an-1+1)2>0,兩邊取以2為底的對(duì)數(shù)可得log2(an+1)=log2(an-1+1)2=2log2(an-1+1),則數(shù)列{log2(an+1)}是以1為首項(xiàng),2為公比的等比數(shù)列,log2(an+1)=2n-1,an=22n-1-1,又an=an-12+2an-1(n≥2),可得an+1=an2+2an(n∈N*),兩邊取倒數(shù)可得1an+1=1an2+2an=1an(an+2)=121an-1an+2,即2an+1=1an-1an+2,因此bn=1an+1+1an+2=1an-1an+1,所以Sn=b1+…+bn=1a1-1an+1=1-122n-1,故答案為1-122n-1.9.(1)解設(shè){an}的公比為q,由題意知q≠1.∴a1(1-q2)1-q=6,　　　　　　　　　　　　　①a1(1-q4)1-q=30,②由②,得a1(1+q2)(1-q2)1-q=30,將①代入,得1+q2=5,q2=4.∵q>0,∴q=2,代入①,得a1=2.∴an=2n.(2)證明由(1)得bn=1(log22n)(log22n+2)=1n(n+2)=121n-1n+2,∴Tn=121-13+1212-14+1213-14+…+121n-1n+2=121+12-1n+1-1n+2<34.10.解(1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.由已知b2+b3=12,得b1(q+q2)=12,而b1=2,所以q2+q-6=0.又因?yàn)閝>0,解得q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.①由S11=11b4,可得a1+5d=16,②聯(lián)立①②,解得a1=1,d=3,由此可得an=3n-2.所以,數(shù)列{an}的通項(xiàng)公式為an=3n-2,數(shù)列{bn}的通項(xiàng)公式為bn=2n.(2)設(shè)數(shù)列{a2nb2n-1}的前n項(xiàng)和為Tn,由a2n=6n-2,b2n-1=2×4n-1,有a2nb2n-1=(3n-1)×4n,故Tn=2×4+5×42+8×43+…+(3n-1)×4n,4Tn=2×42+5×43+8×44+…+(3n-4)×4n+(3n-1)×4n+1,上述兩式相減,得-3Tn=2×4+3×42+3×43+…+3×4n-(3n-1)×4n+1=12×(1-4n)1-4-4-(3n-1)×4n+1=-(3n-2)×4n+1-8.得Tn=3n-23×4n+1+83.所以,數(shù)列{a2nb2n-1}的前n項(xiàng)和為3n-23×4n+1+83.11.解(1)∵bn+1-bn=22an+1-1-22an-1=221-14an-1-22an-1=4an2an-1-22an-1=2(常數(shù)),∴數(shù)列{bn}是等差數(shù)列.∵a1=1,∴b1=2,因此bn=2+(n-1)×2=2n.由bn=22an-1,得an=n+12n.(2)由cn=4ann+1,an=n+12n,得cn=2n,∴cncn+2=4n(n+2)=21n-1n+2,∴Tn=21-13+12-14+13-15+…+1n-1n+2=21+12-1n+1-1n+2<3,依題意要使Tn<1cmcm+1對(duì)于n∈N*恒成立,只需1cmcm+1≥3,即m(m+1)4≥3,解得m≥3或m≤-4.又m為正整數(shù),∴m的最小值為3.6。