高考数学一轮复习单元质检六数列B含解析新人教A版理

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.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.22答案:B2.设an=-n2+9n+10,则数列{an}前n项和最大时n的值为()A.9 B.10 C.9或10 D.12答案:C3.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.60答案: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+10×4.在数列{an}中,a1=1,an+1=2an,Sn为{an}的前n项和.若{Sn+λ}为等比数列,则λ=()A.-1 B.1 C.-2 D.2答案:B解析:由题意,得{an}是等比数列,公比为2,∴Sn=2n-1,Sn+λ=2n-1+λ.∵{Sn+λ}为等比数列,∴-1+λ=0,∴λ=1,故选B.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,∴a解得a∴自上而下取第1,3,9节,这3节的容积之和为a1+a3+a9=3a1+10d=3×1322+10×7666.设a,b∈R,数列{an}满足a1=a,an+1=an2+b,n∈N*,则(A.当b=12时,a10>10 B.当b=14时,a10C.当b=-2时,a10>10 D.当b=-4时,a10>10答案:A解析:考察选项A,a1=a,an+1=an2+b=∵an-122=an2-an+1∵an+1=an2+12>0,an+1≥an-14+∴{an}为递增数列.因此,当a1=0时,a10取到最小值,现对此情况进行估算.显然,a1=0,a2=a12+12=12,a3=a22+12=∴lgan+1>2lgan,∴lga10>2lga9>22·lga8>…>26lga4=lga4∴a10>a464=1+11664=C640+C6411161+C6421162+…+C646411664=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,使三个数成等差数列.若中间项减去6,则三个数成等比数列,则此未知数是.

答案:3或27解析:设此三数为3,a,b,则2a=3+故这个未知数为3或27.8.设Sn是数列{an}的前n项和,且a1=3,当n≥2时,有Sn+Sn-1-2SnSn-1=2nan.则使得S1S2·…·Sm≥2019成立的正整数m的最小值为.

答案:1009解析:∵当n≥2时,Sn+Sn-1-2SnSn-1=2nan,∴Sn+Sn-1-2SnSn-1=2n(Sn-Sn-1),∴2SnSn-1=(2n+1)Sn-1-(2n-1)Sn,易知Sn≠0,∴2n+1S令bn=2n+1Sn,则bn-bn-1∴数列{bn}是以b1=3S1=3a1=1为首项,公差d=2的等差数列,∴bn=2n-∴Sn=2n∴S1S2·…·Sm=3×53×…×2m+12由2m+1≥2019,解得m≥1009,即正整数m的最小值为1009.三、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T解:(1)∵12,an,Sn成等差数列,∴2an=Sn+1当n=1时,2a1=S1+12,即a1=1当n≥2时,an=Sn-Sn-1=2an-2an-1,即ana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),∴1b∴Tn=121-110.(15分)已知数列{an}和{bn}满足a1=2,b1=1,2an+1=an,b1+12b2+13b3+…+1nbn=bn+1(1)求数列{an}和{bn}的通项公式;(2)记数列{anbn}的前n项和为Tn,求Tn.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+…+1nbn=bn+1∴当n=1时,b1=b2-1,故b2=2.当n≥2时,b1+12b2+13b3+…+1n-1bn-1①-②,得1nbn=bn+1-bn得bn+1n+1=(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+120+…+111.(15分)设{an}是等差数列,{bn}是等比数列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求{an}和{bn}的通项公式;(2)设数列{cn}满足c1=1,cn=1,2k<n<①求数列{a2n②求∑i=12naici(n∈解:(1)设等差数列{an}的公差为d,等比数列{bn}的公比为q.依题意得6解得d故an=4+(n-1)×3=3n+1,bn=6×2n-1=3×2n.所以,{an}的通项公式为an=3n+1,{bn}的通项公式为bn=3×2n.(2)①a2n(c2n-1)=a2n(bn-1)=(3×2n+