高考數(shù)學一輪復習單元質(zhì)檢六數(shù)列B含解析新人教A版理

單元質(zhì)檢六數(shù)列(B)(時間: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.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.22答案:B2.設an=-n2+9n+10,則數(shù)列{an}前n項和最大時n的值為()A.9 B.10 C.9或10 D.12答案:C3.公差不為零的等差數(shù)列{an}的前n項和為Sn.若a4是a3與a7的等比中項,S8=16,則S10等于()A.18 B.24 C.30 D.60答案:C解析:設等差數(shù)列{an}的公差為d≠0.由題意,得(a1+3d)2=(a1+2d)(a1+6d),化為2a1+3d=0,①∵S8=16,∴8a1+8×72×聯(lián)立①②解得a1=-32,d=1則S10=10×-32+10×4.在數(shù)列{an}中,a1=1,an+1=2an,Sn為{an}的前n項和.若{Sn+λ}為等比數(shù)列,則λ=()A.-1 B.1 C.-2 D.2答案:B解析:由題意,得{an}是等比數(shù)列,公比為2,∴Sn=2n-1,Sn+λ=2n-1+λ.∵{Sn+λ}為等比數(shù)列,∴-1+λ=0,∴λ=1,故選B.5.《九章算術》中的“竹九節(jié)”問題:現(xiàn)有一根9節(jié)的竹子,自上而下各節(jié)的容積成等差數(shù)列,上面4節(jié)的容積共3升,下面3節(jié)的容積共4升,現(xiàn)自上而下取第1,3,9節(jié),則這3節(jié)的容積之和為()A.133升 B.176升 C.199升 D答案:B解析:設自上而下各節(jié)的容積分別為a1,a2,…,a9,公差為d,∵上面4節(jié)的容積共3升,下面3節(jié)的容積共4升,∴a解得a∴自上而下取第1,3,9節(jié),這3節(jié)的容積之和為a1+a3+a9=3a1+10d=3×1322+10×7666.設a,b∈R,數(shù)列{an}滿足a1=a,an+1=an2+b,n∈N*,則(A.當b=12時,a10>10 B.當b=14時,a10C.當b=-2時,a10>10 D.當b=-4時,a10>10答案:A解析:考察選項A,a1=a,an+1=an2+b=∵an-122=an2-an+1∵an+1=an2+12>0,an+1≥an-14+∴{an}為遞增數(shù)列.因此,當a1=0時,a10取到最小值,現(xiàn)對此情況進行估算.顯然,a1=0,a2=a12+12=12,a3=a22+12=∴l(xiāng)gan+1>2lgan,∴l(xiāng)ga10>2lga9>22·lga8>…>26lga4=lga4∴a10>a464=1+11664=C640+C6411161+C6421162+…+C646411664=1+二、填空題(本大題共2小題,每小題7分,共14分)7.在3和一個未知數(shù)之間填上一個數(shù),使三個數(shù)成等差數(shù)列.若中間項減去6,則三個數(shù)成等比數(shù)列,則此未知數(shù)是.

答案:3或27解析:設此三數(shù)為3,a,b,則2a=3+故這個未知數(shù)為3或27.8.設Sn是數(shù)列{an}的前n項和,且a1=3,當n≥2時,有Sn+Sn-1-2SnSn-1=2nan.則使得S1S2·…·Sm≥2019成立的正整數(shù)m的最小值為.

答案:1009解析:∵當n≥2時,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,則bn-bn-1∴數(shù)列{bn}是以b1=3S1=3a1=1為首項,公差d=2的等差數(shù)列,∴bn=2n-∴Sn=2n∴S1S2·…·Sm=3×53×…×2m+12由2m+1≥2019,解得m≥1009,即正整數(shù)m的最小值為1009.三、解答題(本大題共3小題,共44分)9.(14分)已知數(shù)列{an}的前n項和為Sn,首項為a1,且12,an,Sn成等差數(shù)列(1)求數(shù)列{an}的通項公式;(2)數(shù)列{bn}滿足bn=(log2a2n+1)×(log2a2n+3),求數(shù)列1bn的前n項和T解:(1)∵12,an,Sn成等差數(shù)列,∴2an=Sn+1當n=1時,2a1=S1+12,即a1=1當n≥2時,an=Sn-Sn-1=2an-2an-1,即ana故數(shù)列{an}是首項為12,公比為2的等比數(shù)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)已知數(shù)列{an}和{bn}滿足a1=2,b1=1,2an+1=an,b1+12b2+13b3+…+1nbn=bn+1(1)求數(shù)列{an}和{bn}的通項公式;(2)記數(shù)列{anbn}的前n項和為Tn,求Tn.解:(1)∵2an+1=an,∴{an}是公比為12的等比數(shù)列又a1=2,∴an=2·12∵b1+12b2+13b3+…+1nbn=bn+1∴當n=1時,b1=b2-1,故b2=2.當n≥2時,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則12Tn=120+2以上兩式相減,得12Tn=12-1+120+…+111.(15分)設{an}是等差數(shù)列,{bn}是等比數(shù)列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求{an}和{bn}的通項公式;(2)設數(shù)列{cn}滿足c1=1,cn=1,2k<n<①求數(shù)列{a2n②求∑i=12naici(n∈解:(1)設等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.依題意得6解得d故an=4+(n-1)×3=3n+1,bn=6×2n-1=3×2n.所以,{an}的通項公式為an=3n+1,{bn}的通項公式為bn=3×2n.(2)①a2n(c2n-1)=a2n(bn-1)=(3×2n+