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.已知等差數(shù)列{an}的前n項和為Sn,a3=5,S6=36,則a6=()A.9 B.10 C.11 D.122.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.223.設(shè)an=n2+9n+10,則數(shù)列{an}前n項和最大時n的值為()A.9 B.10C.9或10 D.124.等比數(shù)列{an}的前n項和為Sn,已知S3=a2+5a1,a7=2,則a5=()A.12 B.12 C.2 D5.(2017寧夏銀川一中二模)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.606.(2017遼寧沈陽三模)數(shù)列{an}的前n項和為Sn,a1=1,an+an+1=3×2n1,則S2017=()A.220181 B.22018+1 C.220171 D.22017+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),使三數(shù)成等差數(shù)列,若中間項減去6,則三數(shù)成等比數(shù)列,則此未知數(shù)是.

8.(2017河北石家莊二中模擬)已知數(shù)列{an}滿足:a1=1,an=an-12+2an1(n≥2),若bn=1an+1+1an+2(n∈N*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T10.(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)求an與bn;(2)記數(shù)列{anbn}的前n項和為Tn,求Tn.11.(15分)已知數(shù)列{an}滿足a1=1,an+1=114an,其中n∈(1)設(shè)bn=22an-1,求證:數(shù)列{bn(2)設(shè)cn=4ann+1,數(shù)列{cncn+2}的前n項和為Tn,是否存在正整數(shù)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.故選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項和等于前10項和,它們都最大.4.A解析由條件,得a∴a∴a5=a1q4=132×42=15.C解析設(shè)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+10×6.C解析由a1=1和an+1=3×2n1an,可知數(shù)列{an}唯一確定,并且a2=2,a3=4,a4=8,猜測an=2n1,經(jīng)驗證an=2n1是滿足題意的唯一解.∴S2017=22017-12-7.3或27解析設(shè)此三數(shù)為3,a,b,則2a=3+故這個未知數(shù)為3或27.8.1122n-1解析當(dāng)n≥2時,an+1=an-12+2an1+1=兩邊取以2為底的對數(shù)可得log2(an+1)=log2(an1+1)2=2log2(an1+1),則數(shù)列{log2(an+1)}是以1為首項,2為公比的等比數(shù)列,log2(an+1)=2n1,an=22又an=an-12+2an1(n≥2),可得an+1=an2+2an兩邊取倒數(shù)可得1a即2an+1=1an所以Sn=b1+…+bn=1a1-1an+1=9.解(1)∵12,an,Sn成等差數(shù)列∴2an=Sn+12當(dāng)n=1時,2a1=S1+12,即a1=1當(dāng)n≥2時,an=SnSn1=2an2an1,即ana故數(shù)列{an}是首項為12,公比為2的等比數(shù)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}是公比為12的等比數(shù)列又a1=2,∴an=2·12∵b1+12b2+13b3+…+1=bn+11,①∴當(dāng)n=1時,b1=b21,故b2=2.當(dāng)n≥2時,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則12Tn=120+2以上兩式相減,得12Tn=12-1+故Tn=8n+211.解(1)∵bn+1bn=2=2=4an2a∴數(shù)列{bn}是等差數(shù)列.∵a1=1,∴b1=2,因此bn=2+(n1)×2=2n.由bn=22