高考数学人教A版理科第一轮复习单元测试题单元质检六B

单元质检六数列(B)(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,a3=5,S6=36,则a6=()A.9 B.10 C.11 D.122.在单调递减的等比数列{an}中,若a3=1,a2+a4=52,则a1=(A.2 B.4 C.2 D.223.设an=n2+9n+10,则数列{an}前n项和最大时n的值为()A.9 B.10C.9或10 D.124.等比数列{an}的前n项和为Sn,已知S3=a2+5a1,a7=2,则a5=()A.12 B.12 C.2 D5.(2017宁夏银川一中二模)公差不为零的等差数列{an}的前n项和为Sn.若a4是a3与a7的等比中项,S8=16,则S10等于()A.18 B.24 C.30 D.606.(2017辽宁沈阳三模)数列{an}的前n项和为Sn,a1=1,an+an+1=3×2n1,则S2017=()A.220181 B.22018+1 C.220171 D.22017+1二、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.

8.(2017河北石家庄二中模拟)已知数列{an}满足:a1=1,an=an-12+2an1(n≥2),若bn=1an+1+1an+2(n∈N*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{bn}满足bn=(log2a2n+1)×(log2a2n+3),求数列1bn的前n项和T10.(15分)已知数列{an}和{bn}满足a1=2,b1=1,2an+1=an,b1+12b2+13b3+…+1nbn=bn+(1)求an与bn;(2)记数列{anbn}的前n项和为Tn,求Tn.11.(15分)已知数列{an}满足a1=1,an+1=114an,其中n∈(1)设bn=22an-1,求证:数列{bn(2)设cn=4ann+1,数列{cncn+2}的前n项和为Tn,是否存在正整数m,使得Tn<1cmcm+1答案：1.C解析∵S6=a1+a62×6=(a3+a4)2×6=∴a6=a4+(64)×(75)=11.故选C.2.B解析由已知,得a1q2=1,a1q+a1q3=52∴q+q3q2=52∴q=12(q=2舍去∴a1=4.3.C解析令an≥0,得n29n10≤0,∴1≤n≤10.令an+1≤0,即n27n18≥0,∴n≥9.∴9≤n≤10.∴前9项和等于前10项和,它们都最大.4.A解析由条件,得a∴a∴a5=a1q4=132×42=15.C解析设等差数列{an}的公差为d≠0.由题意得(a1+3d)2=(a1+2d)(a1+6d),化为2a1+3d=0,①∵S8=16,∴8a1+8×72×联立①②解得a1=32,d=1则S10=10×-32+10×6.C解析由a1=1和an+1=3×2n1an,可知数列{an}唯一确定,并且a2=2,a3=4,a4=8,猜测an=2n1,经验证an=2n1是满足题意的唯一解.∴S2017=22017-12-7.3或27解析设此三数为3,a,b,则2a=3+故这个未知数为3或27.8.1122n-1解析当n≥2时,an+1=an-12+2an1+1=两边取以2为底的对数可得log2(an+1)=log2(an1+1)2=2log2(an1+1),则数列{log2(an+1)}是以1为首项,2为公比的等比数列,log2(an+1)=2n1,an=22又an=an-12+2an1(n≥2),可得an+1=an2+2an两边取倒数可得1a即2an+1=1an所以Sn=b1+…+bn=1a1-1an+1=9.解(1)∵12,an,Sn成等差数列∴2an=Sn+12当n=1时,2a1=S1+12,即a1=1当n≥2时,an=SnSn1=2an2an1,即ana故数列{an}是首项为12,公比为2的等比数列,即an=2n2(2)∵bn=(log2a2n+1)×(log2a2n+3)=(log222n+12)×(log222n+32)=(2n1)(2n+1),∴1=12∴Tn=1+=1210.解(1)∵2an+1=an,∴{an}是公比为12的等比数列又a1=2,∴an=2·12∵b1+12b2+13b3+…+1=bn+11,①∴当n=1时,b1=b21,故b2=2.当n≥2时,b1+12b2+13b3+…+1n-1bn①②,得1nbn=bn+1bn得bn+1n+1=b(2)由(1)知anbn=n·12故Tn=12-1+2则12Tn=120+2以上两式相减,得12Tn=12-1+故Tn=8n+211.解(1)∵bn+1bn=2=2=4an2a∴数列{bn}是等差数列.∵a1=1,∴b1=2,因此bn=2+(n1)×2=2n.由bn=22