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.在单调递减的等比数列an中,若a3=1,a2+a4=52,则a1=()A.2B.4C.2D.222.(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.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.124.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
2、中项,S8=16,则S10等于()A.18B.24C.30D.605.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.26.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,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.设Sn是数列an的前n项和,且a1=3,当n2
3、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)已知数列an和bn满足a1=2,b1=1,2an+1=an,b1+12b2+13b3+1nbn=bn+1-1.(1)求数列an和bn的通项公式;(2)记数列anbn的前n项和为Tn,求Tn.11.(15分)(2021河北秦皇岛模拟)
4、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中任选一个,补充到上面的问题中,并给出解答.答案：1.B2.C解析若S35=0,则S35=35(a1+a35)2=0,即a18=0;若a17-a19=0,则-2d=0(d为数列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.C4.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+3
5、d=0,S8=16,8a1+872d=16,联立解得a1=-32,d=1.则S10=10-32+10921=30.5.B解析由题意,得an是等比数列,公比为2,Sn=2n-1,Sn+=2n-1+.Sn+为等比数列,-1+=0,=1,故选B.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
6、,当n1时,an+1an2,lgan+12lgan,lga102lga922lga826lga4=lga464,a10a464=1+11664=C640+C6411161+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+S
7、n-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-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
8、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.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
9、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=12-1+120+12n-2-n2n-1=21-12n1-12-n2n-1,故Tn=8-n+22n-2.11.(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
10、公比的等比数列,bn=12n+1.(2)解选:因为bn=12n+1,所以n+1bn=(n+1)2n+1,则Tn=222+323+(n+1)2n+1,2Tn=223+324+(n+1)2n+2,-,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.
