2021版高考数学（文）导学大一轮人教A广西专用单元质检六　数列（B） WORD版含解析.docx

1、单元质检六数列(B)(时间:45分钟满分:100分)单元质检卷第15页一、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的公差和首项都不等于0,且a2,a4,a8成等比数列,则a1+a5+a9a2+a3=()A.2B.3C.5D.7答案:B解析:设an的公差为d.由题意,得a42=a2a8,(a1+3d)2=(a1+d)(a1+7d),d2=a1d.d0,d=a1,a1+a5+a9a2+a3=15a15a1=3.2.在单调递减的等比数列an中,若a3=1,a2+a4=52,则a1=()A.2B.4C.2D.22答案:B解析:设an的公比为q.由已知,得a1q2=1,a1q+

2、a1q3=52,q+q3q2=52,q2-52q+1=0,q=12(q=2舍去),a1=4.3.在数列an中,a1=1,an+1=2an,Sn为an的前n项和.若Sn+为等比数列,则=()A.-1B.1C.-2D.2答案:B解析:由题意,得an是等比数列,公比为2,Sn=2n-1,Sn+=2n-1+.Sn+为等比数列,-1+=0,=1,故选B.4.已知各项均为正数的等比数列an的前n项和为Sn,若Sn=2,S3n=14,则S4n=()A.80B.26C.30D.16答案:C解析:设各项均为正数的等比数列an的首项为a1,公比为q.Sn=2,S3n=14,a1(1-qn)1-q=2,a1(1-q

3、3n)1-q=14,解得qn=2,a11-q=-2.S4n=a11-q(1-q4n)=-2(1-16)=30.故选C.5.九章算术中的“竹九节”问题:现有一根9节的竹子,自上而下各节的容积成等差数列,上面4节的容积共3升,下面3节的容积共4升,现自上而下取第1,3,9节,则这3节的容积之和为()A.133升B.176升C.199升D.2512升答案:B解析:设自上而下各节的容积分别为a1,a2,a9,公差为d,上面4节的容积共3升,下面3节的容积共4升,a1+a2+a3+a4=4a1+6d=3,a9+a8+a7=3a1+21d=4,解得a1=1322,d=766,自上而下取第1,3,9节,这3

4、节的容积之和为a1+a3+a9=3a1+10d=31322+10766=176(升).6.(2019浙江,10)设a,bR,数列an满足a1=a,an+1=an2+b,nN*,则()A.当b=12时,a1010B.当b=14时,a1010C.当b=-2时,a1010D.当b=-4时,a1010答案:A解析:当b=12时,a2=a12+1212,a3=a22+1234,a4=a32+1217161,当n4时,an+1=an2+12an21,则log1716an+12log1716anlog1716an+12n-1,则an+117162n-1(n4),则a10171626=1+11664=1+64

5、16+646321162+1+4+710,故选A.二、填空题(本大题共2小题,每小题7分,共14分)7.在3和一个未知数之间填上一个数,使三数成等差数列.若中间项减去6,则三数成等比数列,则此未知数是.答案:3或27解析:设此三数为3,a,b,则2a=3+b,(a-6)2=3b,解得a=3,b=3或a=15,b=27.故这个未知数为3或27.8.(2019广东深圳二模)记Sn是数列an的前n项和,且a1=3,当n2时,有Sn+Sn-1-2SnSn-1=2nan.则使得S1S2Sm2 019成立的正整数m的最小值为.答案:1 009解析:Sn+Sn-1-2SnSn-1=2nan,Sn+Sn-1-

6、2SnSn-1=2n(Sn-Sn-1),2SnSn-1=(2n+1)Sn-1-(2n-1)Sn,2n+1Sn-2n-1Sn-1=2.令bn=2n+1Sn,则bn-bn-1=2(n2),数列bn是以b1=3S1=3a1=1为首项,公差d=2的等差数列,bn=2n-1,即2n+1Sn=2n-1,Sn=2n+12n-1,S1S2Sm=3532m+12m-1=2m+1,由2m+12 019,解得m1 009,即正整数m的最小值为1 009.三、解答题(本大题共3小题,共44分)9.(14分)已知数列an的前n项和为Sn,首项为a1,且12,an,Sn成等差数列.(1)求数列an的通项公式;(2)数列b

7、n满足bn=(log2a2n+1)(log2a2n+3),求数列1bn的前n项和Tn.解:(1)12,an,Sn成等差数列,2an=Sn+12.当n=1时,2a1=S1+12,即a1=12;当n2时,an=Sn-Sn-1=2an-2an-1,即anan-1=2,故数列an是首项为12,公比为2的等比数列,即an=2n-2.(2)bn=(log2a2n+1)(log2a2n+3)=(log222n+1-2)(log222n+3-2)=(2n-1)(2n+1),1bn=12n-112n+1=1212n-1-12n+1.Tn=121-13+13-15+12n-1-12n+1=121-12n+1=n2

8、n+1.10.(15分)(2019广东广州高三二模)已知an是递增的等比数列,a2+a3=4,a1a4=3.(1)求数列an的通项公式;(2)令bn=nan,求数列bn的前n项和Sn.解:(1)(方法一)设等比数列an的公比为q,因为a2+a3=4,a1a4=3,所以a1q+a1q2=4,a1a1q3=3,解得a1=9,q=13或a1=13,q=3.因为an是递增的等比数列,所以a1=13,q=3.所以数列an的通项公式为an=3n-2.(方法二)设等比数列an的公比为q,因为a2+a3=4,a1a4=a2a3=3,所以a2,a3是方程x2-4x+3=0的两个根.解得a2=1,a3=3或a2=

9、3,a3=1.因为an是递增的等比数列,所以a2=1,a3=3,则q=3.所以数列an的通项公式为an=3n-2.(2)由(1)知bn=n3n-2.则Sn=13-1+230+331+n3n-2,在式两边同时乘3,得3Sn=130+231+332+n3n-1,-,得-2Sn=3-1+30+31+3n-2-n3n-1,即-2Sn=13(1-3n)1-3-n3n-1,得Sn=14(2n-1)3n-1+112.11.(15分)设an是等比数列,公比大于0,其前n项和为Sn(nN*),bn是等差数列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.(1)求an和bn的通项公式;(2)

10、设数列Sn的前n项和为Tn(nN*),求Tn;证明k=1n(Tk+bk+2)bk(k+1)(k+2)=2n+2n+2-2(nN*).答案:(1)解设等比数列an的公比为q.由a1=1,a3=a2+2,可得q2-q-2=0.因为q0,可得q=2,故an=2n-1.设等差数列bn的公差为d.由a4=b3+b5,可得b1+3d=4.由a5=b4+2b6,可得3b1+13d=16,从而b1=1,d=1,故bn=n.所以,数列an的通项公式为an=2n-1,数列bn的通项公式为bn=n.(2)解由(1),有Sn=1-2n1-2=2n-1,故Tn=k=1n(2k-1)=k=1n2k-n=2(1-2n)1-2-n=2n+1-n-2.证明因为(Tk+bk+2)bk(k+1)(k+2)=(2k+1-k-2+k+2)k(k+1)(k+2)=k2k+1(k+1)(k+2)=2k+2k+2-2k+1k+1,所以,k=1n(Tk+bk+2)bk(k+1)(k+2)=233-222+244-233+2n+2n+2-2n+1n+1=2n+2n+2-2.