廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 單元質(zhì)檢六 數(shù)列（B） 文

單元質(zhì)檢六數(shù)列(B)(時(shí)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的公差和首項(xiàng)都不等于0,且a2,a4,a8成等比數(shù)列,則a1+a5+a9a2+a3=()A.2B.3C.5D.7答案B解析設(shè)an的公差為d.由題意,得a42=a2a8,(a1+3d)2=(a1+d)(a1+7d),d2=a1d.d0,d=a1,a1+a5+a9a2+a3=15a15a1=3.2.在單調(diào)遞減的等比數(shù)列an中,若a3=1,a2+a4=52,則a1=()A.2B.4C.2D.22答案B解析設(shè)an的公比為q.由已知,得a1q2=1,a1q+a1q3=52,q+q3q2=52,q2-52q+1=0,q=12(q=2舍去),a1=4.3.(2018河北唐山期末)在數(shù)列an中,a1=1,an+1=2an,Sn為an的前n項(xiàng)和.若Sn+為等比數(shù)列,則=()A.-1B.1C.-2D.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+=2n-1+.Sn+為等比數(shù)列,-1+=0,=1,故選B.4.(2018陜西西安八校聯(lián)考)設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,若S6>S7>S5,則滿足SnSn+1<0的正整數(shù)n的值為()A.10B.11C.12D.13答案C解析設(shè)an的公差為d.S6>S7>S5,6a1+6×52d>7a1+7×62d>5a1+5×42d,a7<0,a6+a7>0,S13=13(a1+a13)2=13a7<0,S12=12(a1+a12)2=6(a6+a7)>0,滿足SnSn+1<0的正整數(shù)n的值為12,故選C.5.已知各項(xiàng)均為正數(shù)的等比數(shù)列an的前n項(xiàng)和為Sn,若Sn=2,S3n=14,則S4n=()A.80B.26C.30D.16答案C解析設(shè)各項(xiàng)均為正數(shù)的等比數(shù)列an的首項(xiàng)為a1,公比為q.Sn=2,S3n=14,a1(1-qn)1-q=2,a1(1-q3n)1-q=14,解得qn=2,a11-q=-2.S4n=a11-q(1-q4n)=-2×(1-16)=30.故選C.6.(2018河南洛陽(yáng)一模)九章算術(shù)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.2512升答案B解析設(shè)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,a1+a2+a3+a4=4a1+6d=3,a9+a8+a7=3a1+21d=4,解得a1=1322,d=766,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+10×766=176(升).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(gè)未知數(shù)之間填上一個(gè)數(shù),使三數(shù)成等差數(shù)列.若中間項(xiàng)減去6,則三數(shù)成等比數(shù)列,則此未知數(shù)是. 答案3或27解析設(shè)此三數(shù)為3,a,b,則2a=3+b,(a-6)2=3b,解得a=3,b=3或a=15,b=27.故這個(gè)未知數(shù)為3或27.8.已知數(shù)列an滿足:a1=1,an=an-12+2an-1(n2),若bn=1an+1+1an+2(nN*),則數(shù)列bn的前n項(xiàng)和Sn=. 答案1-122n-1解析當(dāng)n2時(shí),an+1=an-12+2an-1+1=(an-1+1)2>0,兩邊取以2為底的對(duì)數(shù)可得log2(an+1)=log2(an-1+1)2=2log2(an-1+1),則數(shù)列l(wèi)og2(an+1)是以1為首項(xiàng),2為公比的等比數(shù)列,log2(an+1)=2n-1,an=22n-1-1,又an=an-12+2an-1(n2),可得an+1=an2+2an(nN*),兩邊取倒數(shù)可得1an+1=1an2+2an=1an(an+2)=121an-1an+2,即2an+1=1an-1an+2,因此bn=1an+1+1an+2=1an-1an+1,所以Sn=b1+bn=1a1-1an+1=1-122n-1,故答案為1-122n-1.三、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(xiàng)和為Sn,首項(xiàng)為a1,且12,an,Sn成等差數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)數(shù)列bn滿足bn=(log2a2n+1)×(log2a2n+3),求數(shù)列1bn的前n項(xiàng)和Tn.解(1)12,an,Sn成等差數(shù)列,2an=Sn+12.當(dāng)n=1時(shí),2a1=S1+12,即a1=12;當(dāng)n2時(shí),an=Sn-Sn-1=2an-2an-1,即anan-1=2,故數(shù)列an是首項(xiàng)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),1bn=12n-1×12n+1=1212n-1-12n+1.Tn=121-13+13-15+12n-1-12n+1=121-12n+1=n2n+1.10.(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.(1)求an與bn;(2)記數(shù)列anbn的前n項(xiàng)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=2·12n-1=12n-2.b1+12b2+13b3+1nbn=bn+1-1,當(dāng)n=1時(shí),b1=b2-1,故b2=2.當(dāng)n2時(shí),b1+12b2+13b3+1n-1bn-1=bn-1,-,得1nbn=bn+1-bn,得bn+1n+1=bnn,故bn=n.(2)由(1)知anbn=n·12n-2=n2n-2.故Tn=12-1+220+n2n-2,則12Tn=120+221+n2n-1.以上兩式相減,得12Tn=12-1+120+12n-2-n2n-1=21-12n1-12-n2n-1,故Tn=8-n+22n-2.11.(15分)設(shè)an是等比數(shù)列,公比大于0,其前n項(xiàng)和為Sn(nN*),bn是等差數(shù)列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.(1)求an和bn的通項(xiàng)公式;(2)設(shè)數(shù)列Sn的前n項(xiàng)和為Tn(nN*),求Tn;證明k=1n(Tk+bk+2)bk(k+1)(k+2)=2n+2n+2-2(nN*).(1)解設(shè)等比數(shù)列an的公比為q.由a1=1,a3=a2+2,可得q2-q-2=0.因?yàn)閝>0,可得q=2,故an=2n-1.設(shè)等差數(shù)列bn的公差為d.由a4=b3+b5,可得b1+3d=4.由a5=b4+2b6,可得3b1+13d=16,從而b1=1,d=1,故bn=n.所以,數(shù)列an的通項(xiàng)公式為an=2n-1,數(shù)列bn的通項(xiàng)公式為bn=n.(2)解由(1),有Sn=1-2n1-2=2n-1,故Tn=k=1n(2k-1)=k=1n2k-n=2×(1-2n)1-2-n=2n+1-n-2.證明因?yàn)?Tk+bk+2)bk(k+1)(k+2)=(2k+1-k-2+k+2)k(k+1)(k+2)=k·2k+1(k+1)(k+2)=2k+2k+2-2k+1k+1,所以,k=1n(Tk+bk+2)bk(k+1)(k+2)=233-222+244-233+2n+2n+2-2n+1n+1=2n+2n+2-2.6