2024年高考指导数学(人教A版理科第一轮复习)课时规范练29　数列的概念

课时规范练29数列的概念基础巩固组1.已知数列12,23,34A.第20项 B.第22项C.第24项 D.第26项2.设Sn为数列{an}的前n项和,且Sn=n2+1,则a5=()A.26 B.19 C.11 D.93.若数列{an}满足a1=1,a2=3,an+1=(2n-λ)an(n=1,2,…),则a3=()A.5 B.9 C.10 D.154.在数列{an}中,an=-2n2+29n+3,则此数列最大项的值是()A.102 B.9658C.9178 D.5.(2022安徽蚌埠三模)若数列{an}满足:a1=1,且an+1=an+3,n为奇数,2aA.19 B.22 C.43 D.466.(2022河南焦作一模)已知数列{an}的前n项和Sn=(-1)nan+12n(n∈N*),则S100=(A.-12100 B.0 C.121007.已知数列{an}的前n项和为Sn,Sn=2n+3,则an=.

8.在数列{an}中,an=n2+kn+4,且对于∀n∈N*,都有an+1>an成立,则实数k的取值范围是.

9.若数列{an}满足a1=35,an+1=2an,0≤an综合提升组10.(2022全国乙,理4)嫦娥二号卫星在完成探月任务后,继续进行深空探测,成为我国第一颗环绕太阳飞行的人造行星.为研究嫦娥二号绕日周期与地球绕日周期的比值,用到数列{bn}:b1=1+1α1,b2=1+1α1+1α2,b3=1+1α1+1αA.b1<b5 B.b3<b8 C.b6<b2 D.b4<b711.数列{an}满足a1=2,an=an+1-1an+1+1,其前n项积为TA.16 B.-16 C.6 D.12.已知数列{an}满足an≠0,2an(1-an+1)-2an+1(1-an)=an-an+1+anan+1,且a1=13,则数列{an}的通项公式an=.创新应用组13.(2022浙江,10)已知数列{an}满足a1=1,an+1=an-13an2(n∈NA.2<100a100<52 B.52<100a100C.3<100a100<72 D.72<100a100答案：课时规范练29数列的概念1.C令nn+1=0.96,解得n=2.D(方法1)依题意Sn=n2+1,当n≥2时,Sn-1=(n-1)2+1=n2-2n+2,an=Sn-Sn-1=2n-1,当n=1时,a1=S1=2,不适合上式.所以an=2所以a5=2×5-1=9.(方法2)a5=S5-S4=9.3.D令n=1,则3=2-λ,即λ=-1,由an+1=(2n+1)an,得a3=5a2=5×3=15.故选D.4.D将an=-2n2+29n+3看作一个二次函数,其对称轴为直线n=294,开口向下,因为n∈N*,所以当n=7时,an取得最大值a7=1085.C由a1=1,得a2=a1+3=4,a3=2a2-1=7,a4=a3+3=10,a5=2a4-1=19,a6=a5+3=22,a7=2a6-1=43.故选C.6.B由题意知S102=a102+12102,所以S102-a102=S101=又因为S101=-a101+12101,所以a101=12102,故S100=S101-a7.5,n=1,2n-1,n≥2因为Sn=2n+3,那么当n=1时当n≥2时,an=Sn-Sn-1=2n+3-(2n-1+3)=2n-1.由于a1=5不满足上式,所以an=58.(-3,+∞)由an+1>an知该数列是递增数列,∵通项公式an=n2+kn+4,∴(n+1)2+k(n+1)+4>n2+kn+4,即k>-1-2n,又n∈N*,∴-1-2n≤-3,∴k>-3.9.35由已知可得,a2=2×35-1=15,a3=2×15=25,a4=2×25=45∴{an}为周期数列且周期为4,∴a2021=a505×4+1=a1=310.D依题意,不妨令ak=1(k=1,2,…),则b1=2,b2=32,b3=53,b4=85,b5=138,b6=2113,b7=3421,b8=5534,所以b1>b5,b3>b8,b6>b2,b11.D由an=an+1-1an+1+1,得an(an+1+1)=an+1-1,即an+1(an-1)=-(an又因为a1=2,则a2=a1+11-a1=2+11-2=-3,a3=a2+11-a2=-3+11-(-3)=-12,a4=a3+11-a3=-12+11-(-12)=13,a5=a4+11-a4=13+11-13=2=a1,…,所以{an}是以4为周期的周期数列,所以a12.1n+2∵an≠0,2an(1-an+1)-2an+1(1-an)=an-an+1+anan+1,∴两边同除以anan+1,得2(1-an+1)an+1−2(1-an)an=1an+1−1an+1,13.B∵a1=1,易得a2=23∈(0,1),依次类推可得an由题意,an+1=an1-13an,即1an∴1即1a2−1a1>1累加可得1an-1>13即1an>13(∴an<3n+2(n≥2),即a100<134,100a100<又1an+1−1an=1∴1a2−1a1=13×1+12,1a3−1a2<13×1+13,