1、单元质检卷六数列(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.(2019北京海淀一模,3)已知等差数列an满足4a3=3a2,则an中一定为零的项是()A.a6B.a8C.a10D.a122.(2019江西九江一模,3)等比数列an中,若a4a5a6=8,且a5与2a6的等差中项为2,则公比q=()A.2B.12C.-2D.-123.已知等差数列an的前n项和为Sn,若2a6=a8+6,则S7=()A.49B.42C.35D.244.(2019湖南湘潭二模)已知数列an为等比数列,首项a1=2,数列bn满足bn=log2an,且b2+b3+b4=9
2、,则a5=()A.8B.16C.32D.645.(2019山西重点中学模拟)13+13+6+13+6+9+13+6+9+30=()A.310B.1033C.35D.20336.(2019黑龙江大庆一中模拟)在各项不为零的等差数列an中,2a2 017-a2 0182+2a2 019=0,数列bn是等比数列,且b2 018=a2 018,则log2(b2 017b2 019)的值为()A.1B.2C.4D.8二、填空题(本大题共2小题,每小题7分,共14分)7.(2019陕西第二次质检,14)公比为2的等比数列an的各项都是正数,且a2a12=16,则log2a15=.8.(2019四川成都高新
3、区模拟,15)已知数列an,若a1+2a2+nan=2n,则数列anan+1的前n项和为.三、解答题(本大题共3小题,共44分)9.(14分)(2019全国2,文18)已知an是各项均为正数的等比数列,a1=2,a3=2a2+16.(1)求an的通项公式;(2)设bn=log2an.求数列bn的前n项和.10.(15分)(2019河南八市联考二,17)已知数列an满足a1a2a3an-1an=n+1(nN*).(1)求数列an的通项公式;(2)若bn=an+1an,求数列bn的前n项和Sn.11.(15分)(2019安徽安庆二模,17)设各项均为正数的数列an的前n项和为Sn,满足对任意的nN
4、*都有an+1+Sn+1=1,又a1=12.(1)求数列an的通项公式;(2)令bn=log2an,求1b1b2+1b2b3+1bnbn+1(nN*).参考答案单元质检卷六数列(A)1.A4a3=3a2,4a1+8d=3a1+3d,则a1+5d=0,即a6=0.2.B根据题意,等比数列an中,若a4a5a6=8,则(a5)3=8,解得a5=2,又由a5与2a6的等差中项为2,则a5+2a6=4,解得a6=1,则q=a6a5=12.故选B.3.B设等差数列an的公差为d,2a6=a8+6,2(a1+5d)=a1+7d+6,a1+3d=6,即a4=6.由等差数列的性质可得a1+a7=2a4.S7=
5、7(a1+a7)2=7a4=42.故选B.4.C设等比数列an的公比为q,已知首项a1=2,所以an=2qn-1,所以bn=log2an=1+(n-1)log2q,所以数列bn是等差数列.因为b2+b3+b4=9,所以3b3=9,解得b3=3,所以a3=23=2q2,解得q2=4,所以a5=224=32.故选C.5.D由题意可知13+6+9+3n=2(3n+3)n=231n-1n+1,原式=231-12+12-13+13-14+110-111=231-111=2033,故选D.6.C由题意a2 017+a2 019=2a2 018,2a2 017-a2 0182+2a2 019=4a2 018
6、-a2 0182=0,由an0,所以a2 018=4,由bn是等比数列,得b2 017b2 019=a2 0182=16,所以log2(b2 017b2 019)=log216=4,故选C.7.6a2a12=a72=16,a7=4,log2a15=log2a7q8=log2a7+log2q8=log24+log2(2)8=2+4=6.故答案为6.8.4nn+1因为a1+2a2+nan=2n,所以a1+2a2+(n-1)an-1=2(n-1),两式相减得nan=2,则an=2n,设数列anan+1的前n项和为Sn,anan+1=2n2n+1=41n-1n+1,则Sn=a1a2+a2a3+a3a4
7、+anan+1=41-12+12-13+1n-1n+1=41-1n+1=4nn+1.9.解 (1)设an的公比为q,由题设得2q2=4q+16,即q2-2q-8=0,解得q=-2(舍去)或q=4.因此an的通项公式为an=24n-1=22n-1.(2)由(1)得bn=(2n-1)log22=2n-1,因此数列bn的前n项和为1+3+2n-1=n2.10.解 (1)数列an满足a1a2a3an-1an=n+1,当n2时,a1a2a3an-1=n,得an=n+1n,当n=1时,a1=2也满足上式,所以数列an的通项公式为an=n+1n.(2)由于an=n+1n,所以bn=an+1an=n+1n+n
8、n+1=1+1n+1-1n+1=2+1n-1n+1,则Sn=2+1-12+2+12-13+2+1n-1n+1=2n+1-1n+1=2n+1-1n+1.11.解 (1)由an+1+Sn+1=1,得an+Sn=1(n2,nN*),-得2an+1-an=0,即an+1=12an(n2,nN*).由a2+S2=a2+(a1+a2)=1,a1=12,得a2=14=12a1,所以an+1=12an(nN*),则数列an是首项和公比都为12的等比数列,因此an=12n(nN*).(2)由an=12n,得bn=log2an=-n,所以1bnbn+1=1n(n+1)=1n-1n+1,所以1b1b2+1b2b3+1bnbn+1=1-12+12-13+1n-1n+1=1-1n+1=nn+1.
