2022年高考数学一轮复习 单元质检6 数列（B）（含解析）新人教A版.docx

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.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.d0,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.22答案:B解析:设an的公比为q.由已知,得a1q2=1,a1q+a1q3=52,q

2、+q3q2=52,q2-52q+1=0,q=12(q=2舍去),a1=4.3.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+为等比数列,-1+=0,=1,故选B.4.设等差数列an的前n项和为Sn,若S6S7S5,则满足SnSn+1S7S5,6a1+652d7a1+762d5a1+542d,a70,S13=13(a1+a13)2=13a70,满足SnSn+10的正整数n的值为12,故选C.5.(2021浙江,10)已知数列an满足

3、a1=1,an+1=an1+an(nN*),记数列an的前n项和为Sn,则()A.32S1003B.3S1004C.4S10092D.92S1005答案:A解析:由a1=1,an+1=an1+an,可知a2=12,0a1+a2=1+12=32.因为an+1=an1+an,所以an+1+an+1an=an,所以an+1=an-an+1an.因为an为递减数列,所以an+an+12an,所以an+1an-an+112(an+an+1)=2(an-an+1).所以S100=a1+a2+a3+a1001+2(a1-a2)+2(a2-a3)+2(a99-a100)=1+2(a1-a2+a2-a3+a99

4、-a100)=1+2(1-a100)=3-2a1003.综上所述,32S1000,两边取以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(n2),可得an+1=an2+2an(nN*),两边取倒数可得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-

5、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)=(log222

6、n+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=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

7、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.(15分)已知an为等差数列,bn为等比数列,a1=b1=1,a5=5(a4-a3),b5=4(b4-b3).(1)求an和bn的通项公式;(2)记an的前n项和为Sn,求证:SnSn+2Sn+12(nN*

8、);(3)对任意的正整数n,设cn=(3an-2)bnanan+2,n为奇数,an-1bn+1,n为偶数.求数列cn的前2n项和.答案:(1)解设等差数列an的公差为d,等比数列bn的公比为q.由a1=1,a5=5(a4-a3),可得d=1,从而an的通项公式为an=n.由b1=1,b5=4(b4-b3),又q0,可得q2-4q+4=0,解得q=2,从而bn的通项公式为bn=2n-1.(2)证明由(1)可得Sn=n(n+1)2,故SnSn+2=14n(n+1)(n+2)(n+3),Sn+12=14(n+1)2(n+2)2,从而SnSn+2-Sn+12=-12(n+1)(n+2)0,所以SnSn

9、+2Sn+12.(3)解当n为奇数时,cn=(3an-2)bnanan+2=(3n-2)2n-1n(n+2)=2n+1n+2-2n-1n;当n为偶数时,cn=an-1bn+1=n-12n.对任意的正整数n,有k=1nc2k-1=k=1n22k2k+1-22k-22k-1=22n2n+1-1,和k=1nc2k=k=1n2k-14k=14+342+543+2n-14n.由得14k=1nc2k=142+343+2n-34n+2n-14n+1.由得34k=1nc2k=14+242+24n-2n-14n+1=241-14n1-14-14-2n-14n+1,从而得k=1nc2k=59-6n+594n.因此,k=12nck=k=1nc2k-1+k=1nc2k=4n2n+1-6n+594n-49.所以,数列cn的前2n项和为4n2n+1-6n+594n-49.