2024届一轮复习人教A版 数列的概念与简单表示法 作业

作业29数列的概念与简单表示法(时间:120分钟满分:150分)1.设数列{an}的前n项和为Sn,且Sn=2an-1(n∈N*),则a6=()A.12 B.13 C.16 D.322.(2023广东汕头三模)已知数列{an}中,a1=-14,当n>1时,an=1-1an-1,则aA.-14 B.C.5 D.-43.(2023青海西宁二模)数列{an}满足a1+2a2+3a3+…+nan=(n-1)·2n+1,则a7=()A.64 B.128 C.256 D.5124.(多选)设Sn是数列{an}的前n项和,a1=1,an+1+SnSn+1=0,则下列说法正确的有()A.数列{an}的前n项和为Sn=1nB.数列1Sn为递增数列C.数列{an}的通项公式为an=-1nD.数列{an}的最大项为a15.已知数列{an}的前n项和公式为Sn=n2-2n+1,则其通项公式为an=.

6.(2023湖北黄冈模拟)已知数列{an}的通项公式为an=n2+λn,若数列{an}为递增数列,则λ的取值范围是.

综合提升组7.(2023江苏盐城模拟)已知数列{an}满足a1=1,nan+1=2(n+1)an,设bn=ann,则数列{bn}的前6项和为(A.127 B.255 C.31 D.638.已知数列{an}的前n项和为Sn,Sn=n2an,a1=1,则Sn=()A.2nn+1 BC.n22n9.在数列{an}中,1a1+12a2+13a3+…+110.已知数列{an}的前n项和为Sn,若a1=1,an>0,8Sn2=an+1(2Sn+an+1),则S10a创新应用组11.已知数列{an}的前n项和为Sn,a1=1,a2=2,an=3an-1+4an-2(n≥3),则S10=()A.410-15C.410-1 D.411-112.在数列{an}中,a1+2a2+3a3+…+nan=2n,则a1a24+a2aA.710 B.13C.95 D.13.已知数列{an}满足a1=1,an=a1+12a2+13a3+…+1n-1an-1(n>1),则数列{an作业29数列的概念与简单表示法1.D解析:当n=1时,a1=S1=2a1-1,可得a1=p1.当n≥2时,an=Sn-Sn-1=2(an-an-1),即an=2an-p1.故数列{an}是首项为1,公比为2的等比数列,其通项公式为an=2n-1,a6=32.故选D.2.B解析:由题意得a2=1-1a1=5,a3=1-1a2=45,a4=1-1a3=-14,则数列{an}的周期为3,故a3.A解析:当n≥2时,由a1+2a2+3a3+…+nan=(n-1)·2n+1,①得a1+2a2+3a3+…+(n-1)an-1=(n-2)·2n-1+1,②①-②,得nan=[(n-1)·2n+1]-[(n-2)·2n-1+1]=n·2n-1(n≥2),所以an=2n-1(n≥2),则a7=64.4.ABD解析:由an+1+SnSn+1=0,得Sn+1-Sn=-SnSn+p1.易知Sn≠0,∴1Sn-1Sn+1=-1,即1Sn+1-1Sn=p1.又1S1=1a1=1,∴数列1Sn为以1为首项,1为公差的等差数列,则1Sn=1+(n-1)×1=n,可得Sn=1n,故选项A、选项B正确;当n≥2时,an=Sn-Sn-1=1n-1n-1=n-15.0,n=1,2n-3,n≥2解析:由题意,数列{an}的前n项和公式为Sn=n2-2n+p1.当n≥2时,an=Sn-Sn-1=n2-2n+1-(n-1)2+2(n-1)-1=2n-3.又当n=1时,a1=S1=12-2×1+1=0,6.λ>-3解析:依题意对于∀n∈N*,都有an+1-an>0,即(n+1)2+λ(n+1)-n2-λn>0,解得λ>-2n-p1.又(-2n-1)max=-3,∴λ>-3.7.D解析:由an+1n+1=2×ann及bn=ann,得bn+1=2bn,又b1=a11=1,所以数列{bn}是首项为1,公比为2的等比数列,于是数列{bn}8.A解析:S2=22a2,即1+a2=4a2,所以a2=13.因为Sn=n2an,所以Sn+1=(n+1)2an+p1.两式作差得Sn+1-Sn=(n+1)2an+1-n2an,即an+1=(n+1)2an+1-n2an,即(n+2)an+1=nan,所以an+1an=nn+2,即anan-1=n-1n+1(n≥2),则an=所以Sn=21-12+12-13+…+1n-1n+1=219.2解析:由题得1a1+12a21a1+12a2+1①-②得1n所以1an=3n4n2-1,所以an=4n2-13n=43n-13n(n≥2).因为n=1也满足,所以an=43n-13n(n∈N*),所以数列{an}为递增数列,10.32解析:由8Sn2=an+1(2Sn+an+1),整理得8Sn2-2an+1Sn-an+12=0,故(4Sn+an+1)(2Sn-an+1)=0.因为an>0,所以4Sn+an+1>0,故2Sn=an+1=Sn+1-Sn,整理得Sn+1=3Sn,故数列{Sn}是以1为首项,3为公比的等比数列,故Sn1p1.A解析:因为an=3an-1+4an-2(n≥3),所以an+an-1=4(an-1+an-2).又a1+a2=3≠0,所以an+an-1an-1+an-2=4(n≥3),所以{an+an+1}是公比为4,首项为3的等比数列,则数列{a2n+a2n-1}也是等比数列,公比为42=16,首项为3,所以S10=(a1+a2)+(a3+a4)12.A解析:已知a1+2a2+3a3+…+nan=2n,则当n≥2时,a1+2a2+3a3+…+(n-1)an-1=2n-1,两式相减得nan=2n-1,即an=2n-1n,n≥2,当n=1时,a1=21=2,不符合,则an=2,n=1,2n-1n,n≥2.当k≥2时,akak+14k=2k-1·2k(13.an=1,