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}的公差和首项都不等于0,且a2,a4,a8成等比数列,则a1+a5+a9a2+a3=()A.2B.3C.5D.7答案B解析设{an}的公差为d.由题意,得a42=a2a8,∴(a1+3d)2=(a1+d)(a1+7d),∴d2=a1d.∵d≠0,∴d=a1,∴a1+a5+a9a2+a3=15a15a1=3.2.在单调递减的等比数列{an}中,若a3=1,a2+a4=52,则a1=()A.2B.4C.❑√2D.2❑√2答案B解析设{an}的公比为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.-1B.1C.-2D.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.4.(2018陕西西安八校联考)设等差数列{an}的前n项和为Sn,若S6>S7>S5,则满足SnSn+1<0的正整数n的值为()A.10B.11C.12D.13答案C解析设{an}的公差为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,∴满足SnSn+1<0的正整数n的值为12,故选C.5.已知各项均为正数的等比数列{an}的前n项和为Sn,若Sn=2,S3n=14,则S4n=()A.80B.26C.30D.16答案C解析设各项均为正数的等比数列{an}的首项为a1,公比为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.故选C.6.(2018河南洛阳一模)《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.2512升答案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,∴{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节,这3节的容积之和为a1+a3+a9=3a1+10d=3×1322+10×766=176(升).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.答案3或27解析设此三数为3,a,b,则{2a=3+b,(a-6)2=3b,解得{a=3,b=3或{a=15,b=27.故这个未知数为3或27.8.已知数列{an}满足:a1=1,an=an-12+2an-1(n≥2),若bn=1an+1+1an+2(n∈N*),则数列{bn}的前n项和Sn=.答案1-122n-1解析当n≥2时,an+1=an-12+2an-1+1=(an-1+1)2>0,两边取以2为底的对数可得log2(an+1)=log2(an-1+1)2=2log2(an-1+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)=2n-1,an=22n-1-1,又an=an-12+2an-1(n≥2),可得an+1=an2+2an(n∈N*),两边取倒数可得1an+1=1an2+2an=1an(an+2)=12(1an-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,故答案为1-122n-1.三、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Tn.解(1)∵12,an,Sn成等差数列,∴2an=Sn+12.当n=1时,2a1=S1+12,即a1=12;当n≥2时,an=Sn-Sn-1=2an-2an-1,即anan-1=2,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),∴1bn=12n-1×12n+1=12(12n-1-12n+1).∴Tn=12[(1-13)+(13-15)+…+(12n-1-12n+1)]=12(1-12n+1)=n2n+1.10.(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.(1)求an与bn;(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=2·12n-1=12n-2.∵b1+12b2+13b3+…+1nbn=bn+1-1,①∴当n=1时,b1=b2-1,故b2=2.当n≥2时,b1+12b2+13b3+…+1n-1bn-1=bn-1,②①-②,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,则12Tn=120+221+…+n2n-1.以上两式相减,得12Tn=12-1+120+…+12n-2−n2n-1=2(1-12n)1-12−n2n-1,故Tn=8-n+22n-2.11.(15分)设{an}是等比数列,公比大于0,其前n项和为Sn(n∈N*),{bn}是等差数列.已知a1=1,a3=a2+2,a4=b3+b5,a5=b4+2b6.(1)求{an}和{bn}的通项公式;(2)设数列{Sn}的前n项和为Tn(n∈N*),①求Tn;②证明∑k=1n(Tk+bk+2)bk(k+1)(k+2)=2n+2n+2-2(n∈N*).(1)解设等比数列{an}的公比为q.由a1=1,a3=a2+2,可得q2-q-2=0.因为q>0,可得q=2,故an=2n-1.设等差数列{bn}的公差为d.由a4=b3+b5,可得b1+3d=4.由a5=b4+2b6,可得3b1+13d=16,从而b1=1,d=1,故bn=n.所以,数列{an}的通项公式为an=2n-1,数列{bn}的通项公式为bn=n.(2)①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-2.②证明因为(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.
