2024屆一輪復(fù)習(xí)人教A版 數(shù)列的概念與簡(jiǎn)單表示法 作業(yè)

作業(yè)29數(shù)列的概念與簡(jiǎn)單表示法(時(shí)間:120分鐘滿分:150分)1.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-1(n∈N*),則a6=()A.12 B.13 C.16 D.322.(2023廣東汕頭三模)已知數(shù)列{an}中,a1=-14,當(dāng)n>1時(shí),an=1-1an-1,則aA.-14 B.C.5 D.-43.(2023青海西寧二模)數(shù)列{an}滿足a1+2a2+3a3+…+nan=(n-1)·2n+1,則a7=()A.64 B.128 C.256 D.5124.(多選)設(shè)Sn是數(shù)列{an}的前n項(xiàng)和,a1=1,an+1+SnSn+1=0,則下列說(shuō)法正確的有()A.數(shù)列{an}的前n項(xiàng)和為Sn=1nB.數(shù)列1Sn為遞增數(shù)列C.數(shù)列{an}的通項(xiàng)公式為an=-1nD.數(shù)列{an}的最大項(xiàng)為a15.已知數(shù)列{an}的前n項(xiàng)和公式為Sn=n2-2n+1,則其通項(xiàng)公式為an=.

6.(2023湖北黃岡模擬)已知數(shù)列{an}的通項(xiàng)公式為an=n2+λn,若數(shù)列{an}為遞增數(shù)列,則λ的取值范圍是.

綜合提升組7.(2023江蘇鹽城模擬)已知數(shù)列{an}滿足a1=1,nan+1=2(n+1)an,設(shè)bn=ann,則數(shù)列{bn}的前6項(xiàng)和為(A.127 B.255 C.31 D.638.已知數(shù)列{an}的前n項(xiàng)和為Sn,Sn=n2an,a1=1,則Sn=()A.2nn+1 BC.n22n9.在數(shù)列{an}中,1a1+12a2+13a3+…+110.已知數(shù)列{an}的前n項(xiàng)和為Sn,若a1=1,an>0,8Sn2=an+1(2Sn+an+1),則S10a創(chuàng)新應(yīng)用組11.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,a2=2,an=3an-1+4an-2(n≥3),則S10=()A.410-15C.410-1 D.411-112.在數(shù)列{an}中,a1+2a2+3a3+…+nan=2n,則a1a24+a2aA.710 B.13C.95 D.13.已知數(shù)列{an}滿足a1=1,an=a1+12a2+13a3+…+1n-1an-1(n>1),則數(shù)列{an作業(yè)29數(shù)列的概念與簡(jiǎn)單表示法1.D解析:當(dāng)n=1時(shí),a1=S1=2a1-1,可得a1=p1.當(dāng)n≥2時(shí),an=Sn-Sn-1=2(an-an-1),即an=2an-p1.故數(shù)列{an}是首項(xiàng)為1,公比為2的等比數(shù)列,其通項(xiàng)公式為an=2n-1,a6=32.故選D.2.B解析:由題意得a2=1-1a1=5,a3=1-1a2=45,a4=1-1a3=-14,則數(shù)列{an}的周期為3,故a3.A解析:當(dāng)n≥2時(shí),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),則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,∴數(shù)列1Sn為以1為首項(xiàng),1為公差的等差數(shù)列,則1Sn=1+(n-1)×1=n,可得Sn=1n,故選項(xiàng)A、選項(xiàng)B正確;當(dāng)n≥2時(shí),an=Sn-Sn-1=1n-1n-1=n-15.0,n=1,2n-3,n≥2解析:由題意,數(shù)列{an}的前n項(xiàng)和公式為Sn=n2-2n+p1.當(dāng)n≥2時(shí),an=Sn-Sn-1=n2-2n+1-(n-1)2+2(n-1)-1=2n-3.又當(dāng)n=1時(shí),a1=S1=12-2×1+1=0,6.λ>-3解析:依題意對(duì)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,所以數(shù)列{bn}是首項(xiàng)為1,公比為2的等比數(shù)列,于是數(shù)列{bn}8.A解析:S2=22a2,即1+a2=4a2,所以a2=13.因?yàn)镾n=n2an,所以Sn+1=(n+1)2an+p1.兩式作差得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),則an=所以Sn=21-12+12-13+…+1n-1n+1=219.2解析:由題得1a1+12a21a1+12a2+1①-②得1n所以1an=3n4n2-1,所以an=4n2-13n=43n-13n(n≥2).因?yàn)閚=1也滿足,所以an=43n-13n(n∈N*),所以數(shù)列{an}為遞增數(shù)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?yàn)閍n>0,所以4Sn+an+1>0,故2Sn=an+1=Sn+1-Sn,整理得Sn+1=3Sn,故數(shù)列{Sn}是以1為首項(xiàng),3為公比的等比數(shù)列,故Sn1p1.A解析:因?yàn)閍n=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}是公比為4,首項(xiàng)為3的等比數(shù)列,則數(shù)列{a2n+a2n-1}也是等比數(shù)列,公比為42=16,首項(xiàng)為3,所以S10=(a1+a2)+(a3+a4)12.A解析:已知a1+2a2+3a3+…+nan=2n,則當(dāng)n≥2時(shí),a1+2a2+3a3+…+(n-1)an-1=2n-1,兩式相減得nan=2n-1,即an=2n-1n,n≥2,當(dāng)n=1時(shí),a1=21=2,不符合,則an=2,n=1,2n-1n,n≥2.當(dāng)k≥2時(shí),akak+14k=2k-1·2k(13.an=1,