2024年高考指導(dǎo)數(shù)學(xué)(人教A版理科第一輪復(fù)習(xí))課時(shí)規(guī)范練29　數(shù)列的概念

課時(shí)規(guī)范練29數(shù)列的概念基礎(chǔ)鞏固組1.已知數(shù)列12,23,34A.第20項(xiàng) B.第22項(xiàng)C.第24項(xiàng) D.第26項(xiàng)2.設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,且Sn=n2+1,則a5=()A.26 B.19 C.11 D.93.若數(shù)列{an}滿足a1=1,a2=3,an+1=(2n-λ)an(n=1,2,…),則a3=()A.5 B.9 C.10 D.154.在數(shù)列{an}中,an=-2n2+29n+3,則此數(shù)列最大項(xiàng)的值是()A.102 B.9658C.9178 D.5.(2022安徽蚌埠三模)若數(shù)列{an}滿足:a1=1,且an+1=an+3,n為奇數(shù),2aA.19 B.22 C.43 D.466.(2022河南焦作一模)已知數(shù)列{an}的前n項(xiàng)和Sn=(-1)nan+12n(n∈N*),則S100=(A.-12100 B.0 C.121007.已知數(shù)列{an}的前n項(xiàng)和為Sn,Sn=2n+3,則an=.

8.在數(shù)列{an}中,an=n2+kn+4,且對(duì)于?n∈N*,都有an+1>an成立,則實(shí)數(shù)k的取值范圍是.

9.若數(shù)列{an}滿足a1=35,an+1=2an,0≤an綜合提升組10.(2022全國乙,理4)嫦娥二號(hào)衛(wèi)星在完成探月任務(wù)后,繼續(xù)進(jìn)行深空探測,成為我國第一顆環(huán)繞太陽飛行的人造行星.為研究嫦娥二號(hào)繞日周期與地球繞日周期的比值,用到數(shù)列{bn}:b1=1+1α1,b2=1+1α1+1α2,b3=1+1α1+1αA.b1<b5 B.b3<b8 C.b6<b2 D.b4<b711.數(shù)列{an}滿足a1=2,an=an+1-1an+1+1,其前n項(xiàng)積為TA.16 B.-16 C.6 D.12.已知數(shù)列{an}滿足an≠0,2an(1-an+1)-2an+1(1-an)=an-an+1+anan+1,且a1=13,則數(shù)列{an}的通項(xiàng)公式an=.創(chuàng)新應(yīng)用組13.(2022浙江,10)已知數(shù)列{an}滿足a1=1,an+1=an-13an2(n∈NA.2<100a100<52 B.52<100a100C.3<100a100<72 D.72<100a100答案：課時(shí)規(guī)范練29數(shù)列的概念1.C令nn+1=0.96,解得n=2.D(方法1)依題意Sn=n2+1,當(dāng)n≥2時(shí),Sn-1=(n-1)2+1=n2-2n+2,an=Sn-Sn-1=2n-1,當(dāng)n=1時(shí),a1=S1=2,不適合上式.所以an=2所以a5=2×5-1=9.(方法2)a5=S5-S4=9.3.D令n=1,則3=2-λ,即λ=-1,由an+1=(2n+1)an,得a3=5a2=5×3=15.故選D.4.D將an=-2n2+29n+3看作一個(gè)二次函數(shù),其對(duì)稱軸為直線n=294,開口向下,因?yàn)閚∈N*,所以當(dāng)n=7時(shí),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.故選C.6.B由題意知S102=a102+12102,所以S102-a102=S101=又因?yàn)镾101=-a101+12101,所以a101=12102,故S100=S101-a7.5,n=1,2n-1,n≥2因?yàn)镾n=2n+3,那么當(dāng)n=1時(shí)當(dāng)n≥2時(shí),an=Sn-Sn-1=2n+3-(2n-1+3)=2n-1.由于a1=5不滿足上式,所以an=58.(-3,+∞)由an+1>an知該數(shù)列是遞增數(shù)列,∵通項(xiàng)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}為周期數(shù)列且周期為4,∴a2021=a505×4+1=a1=310.D依題意,不妨令ak=1(k=1,2,…),則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?yàn)閍1=2,則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為周期的周期數(shù)列,所以a12.1n+2∵an≠0,2an(1-an+1)-2an+1(1-an)=an-an+1+anan+1,∴兩邊同除以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),依次類推可得an由題意,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,