1、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.72.(2021云南昭通模拟)已知数列an是等差数列,其前n项和为Sn,有下列四个命题:甲:a18=0;乙:S35=0;丙:a17-a19=0;丁:S19-S16=0.如果只有一个是假命题,则该命题是()A.甲B.乙C.丙D.丁3.记等差数列an的前n项和为Sn.若S10=95,a8=17,则()A.an=5n-23B.Sn=2n2-212nC.an=4n-15D.
2、Sn=3n2-11n24.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.165.九章算术中的“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6.设a,bR,数列an满足a1=a,an+1=an2+b,nN*,则()A.当b=12时,a1010B.当b=14时,a1010C.当b=-2时,a1010D.当b=-4时,a1010二、填空题(本大题共2小题,每小题7分,
3、共14分)7.(2021江苏镇江信息考试)各项均为正数的等比数列an,其公比q1,且a3a7=4,请写出一个符合条件的通项公式an=.8.记Sn是数列an的前n项和,且a1=3,当n2时,有Sn+Sn-1-2SnSn-1=2nan.则使得S1S2Sm2 019成立的正整数m的最小值为.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.10.(15分)(2021山东淄博一模)将n2(nN*)个正数排
4、成n行n列:a11a12a13a14a1na21a22a23a24a2na31a32a33a34a3na41a42a43a44a4nan1an2an3an4ann其中每一行的数成等差数列,每一列的数成等比数列,并且各列的公比都相等,若a11=1,a13a23a33=1,a32+a33+a34=32.(1)求a1n;(2)设Sn=a11+a22+a33+ann,求Sn.11.(15分)(2021河北秦皇岛模拟)已知数列an满足2an+1=an+1,a1=54,bn=an-1.(1)求证:数列bn是等比数列;(2)求数列的前n项和Tn.从条件n+1bn,n+bn,4log2bnlog2bn+1中任
5、选一个,补充到上面的问题中,并给出解答.答案：1.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.d0,d=a1,a1+a5+a9a2+a3=15a15a1=3.2.C解析若S35=0,则S35=35(a1+a35)2=0,即a18=0;若a17-a19=0,则-2d=0(d为an的公差),即d=0;若S19-S16=a17+a18+a19=0,则a18=0.又因为只有一个是假命题,所以丙是假命题.3.D解析设等差数列an的公差为d,S10=95,a8=17,10a1+45d=95,a1+7d=17,解得a1=-4,d=3,an=
6、-4+3(n-1)=3n-7,Sn=-4n+n(n-1)23=3n2-11n2.故选D.4.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.5.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
7、+a3+a9=3a1+10d=31322+10766=176(升).6.A解析考察选项A,a1=a,an+1=an2+b=an2+12,an-122=an2-an+140,an2an-14.an+1=an2+120,an+1an-14+12=an+14an,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=34,a4=a32+12=1716,当n1时,an+1an2,lgan+12lgan,lga102lga922lga826lga4=lga464,a10a464=1+11664=C640+C6411161+
8、C6421162+C646411664=1+64116+646321162+11664=1+4+7.875+11664=12.875+1166410,因此符合题意,故选A.7.2n-4(只要an为正项等比数列(不为常数列),且a5=2即可)解析因为an为正项等比数列,所以a3a7=a52=4,所以a5=2.又因为q1,不妨令q=2,所以an=a1qn-1=a5qn-5=22n-5=2n-4.8.1 009解析当n2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Sn0,2n+1Sn
9、-2n-1Sn-1=2.令bn=2n+1Sn,则bn-bn-1=2(n2),数列bn是以b1=3S1=3a1=1为首项,公差d=2的等差数列,bn=2n-1,即2n+1Sn=2n-1,Sn=2n+12n-1,S1S2Sm=3532m+12m-1=2m+1,由2m+12019,解得m1009,即正整数m的最小值为1009.9.解(1)12,an,Sn成等差数列,2an=Sn+12.当n=1时,2a1=S1+12,即a1=12;当n2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)(l
10、og2a2n+3)=(log222n+1-2)(log222n+3-2)=(2n-1)(2n+1),1bn=12n-112n+1=1212n-1-12n+1.Tn=121-13+13-15+12n-1-12n+1=121-12n+1=n2n+1.10.解(1)设第一行数的公差为d,各列的公比为q.由题意可知a13a23a33=a233=1,解得a23=1,由a32+a33+a34=3a33=32,解得a33=12,则q=a33a23=12.由a23=a13q=(a11+2d)q=(1+2d)12=1,解得d=12,因此a1n=a11+(n-1)d=1+n-12=n+12.(2)由ann=a1n
11、qn-1=n+1212n-1=n+12n,可得Sn=221+322+423+n+12n,两边同时乘以12可得12Sn=222+323+n2n+n+12n+1,上述两式相减可得12Sn=1+122+123+12n-n+12n+1=1+1221-12n-11-12-n+12n+1=32-n+32n+1,因此,Sn=3-n+32n.11.(1)证明因为2an+1=an+1,所以2an+1-2=an-1.又因为bn=an-1,所以2bn+1=bn,bn+1bn=12.因为b1=a1-1=14,所以数列bn是以14为首项,12为公比的等比数列,bn=12n+1.(2)解选:因为bn=12n+1,所以n+
12、1bn=(n+1)2n+1,则Tn=222+323+(n+1)2n+1,2Tn=223+324+(n+1)2n+2,两式相减,得Tn=-222-(23+24+2n+1)+(n+1)2n+2=-23-23(1-2n-1)1-2+(n+1)2n+2=n2n+2.故Tn=n2n+2.选:因为bn=12n+1,所以n+bn=n+12n+1,则Tn=1+14+2+18+3+116+n+12n+1=(1+2+3+n)+14+18+116+12n+1=12n(n+1)+141-12n1-12=n22+n2+12-12n+1,故Tn=n22+n2+12-12n+1.选:因为bn=12n+1,所以4log2bnlog2bn+1=41n+1-1n+2,则Tn=412-13+13-14+1n-1n+1+1n+1-1n+2=412-1n+2=2nn+2,故Tn=2nn+2.