1、单元质检六数列(B)(时间:45分钟满分:100分)单元质检卷第12页一、选择题(本大题共6小题,每小题7分,共42分)1.在单调递减的等比数列an中,若a3=1,a2+a4=52,则a1=()A.2B.4C.2D.22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=4.2.设an=-n2+9n+10,则数列an前n项和最大时n的值为()A.9B.10C.9或10D.12答案:C解析:令an0,得n2-9n-100,1n10.令an+10,即n2-7n-180,n9.9n10.前9项和等于前
2、10项和,它们都最大.3.公差不为零的等差数列an的前n项和为Sn.若a4是a3与a7的等比中项,S8=16,则S10等于()A.18B.24C.30D.60答案:C解析:设等差数列an的公差为d0.由题意,得(a1+3d)2=(a1+2d)(a1+6d),化为2a1+3d=0,S8=16,8a1+872d=16,联立解得a1=-32,d=1.则S10=10-32+10921=30.4.在数列an中,a1=1,an+1=2an,Sn为an的前n项和.若Sn+为等比数列,则=()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+=2n-1+.Sn
3、+为等比数列,-1+=0,=1,故选B.5.九章算术中的“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=31322+10766=176
4、(升).6.(2019浙江,10)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答案:A解析:当b=12时,a2=a12+1212,a3=a22+1234,a4=a32+1217161,当n4时,an+1=an2+12an21,则log1716an+12log1716anlog1716an+12n-1,则an+117162n-1(n4),则a10171626=1+11664=1+6416+646321162+1+4+710,故选A.二、填空题(本大题共2小题,
5、每小题7分,共14分)7.在3和一个未知数之间填上一个数,使三个数成等差数列.若中间项减去6,则三个数成等比数列,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.(2019广东深圳二模)设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.答案:1 009解析:Sn+Sn-1-2SnSn-1=2nan,Sn+Sn-1-2SnSn-1=2n(Sn-Sn-1),2SnSn-1=(2n+1)Sn
6、-1-(2n-1)Sn,2n+1Sn-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1009.三、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
7、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)(log2a2n+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.(15分)已知数列an和bn满足a1=2,b1=1,2an+1=
8、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=212n-1=12n-2.b1+12b2+13b3+1nbn=bn+1-1,当n=1时,b1=b2-1,故b2=2.当n2时,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=n12n-2=n2n-2.故Tn=12-1+220+n2n-2,则12Tn=120+221+n2n-1.以上两式相减,得12Tn
9、=12-1+120+12n-2-n2n-1=21-12n1-12-n2n-1,故Tn=8-n+22n-2.11.(15分)(2019天津,理19)设an是等差数列,bn是等比数列.已知a1=4,b1=6,b2=2a2-2,b3=2a3+4.(1)求an和bn的通项公式;(2)设数列cn满足c1=1,cn=1,2kn2k+1,bk,n=2k,其中kN*.求数列a2n(c2n-1)的通项公式;求i=12naici(nN*).解:(1)设等差数列an的公差为d,等比数列bn的公比为q.依题意得6q=6+2d,6q2=12+4d,解得d=3,q=2,故an=4+(n-1)3=3n+1,bn=62n-1=32n.所以,an的通项公式为an=3n+1,bn的通项公式为bn=32n.(2)a2n(c2n-1)=a2n(bn-1)=(32n+1)(32n-1)=94n-1.所以,数列a2n(c2n-1)的通项公式为a2n(c2n-1)=94n-1.i=12naici=i=12nai+ai(ci-1)=i=12nai+i=1na2i(c2i-1)=2n4+2n(2n-1)23+i=1n(94i-1)=(322n-1+52n-1)+94(1-4n)1-4-n=2722n-1+52n-1-n-12(nN*).
