2025版高考數(shù)學(xué)復(fù)習(xí)第五單元第29講數(shù)列求和練習(xí)文含解析新人教A版

PAGEPAGE1第29講數(shù)列求和1.[2024·南寧模擬]已知{an}是公差為1的等差數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和,若S8=4S4,則a10= ()A.172 B.192 C.10 D2.[2024·濰坊二模]設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,若Sn=-n2-n,則數(shù)列2(n+1)anA.3940 B.-3940 C.4041 3.[2024·山西榆社中學(xué)模擬]設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,已知a1=12,n+1an+1=nan+2n,A.2-492100 B.2C.2-512100 D.24.[2024·成都第七中學(xué)一診]已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a9=12a12+6,a2=4,則數(shù)列1Sn的前10項(xiàng)和為A.1112 B.1011 C.910 5.[2024·江淮十校三聯(lián)]若數(shù)列{an}的通項(xiàng)公式是an=(-1)n+1·(3n-2),則a1+a2+…+a2024= ()A.-3027 B.3027 C.-3030 D.30306.[2024·株洲一模]已知數(shù)列{an}的前n項(xiàng)和為Sn,且a1=1,2Sn=an+1,則Sn= ()A.2n-1 B.2n-1C.3n-1 D.12(3n-7.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,若S3=7,S6=63,則數(shù)列{nan}的前n項(xiàng)和為 ()A.-3+(n+1)×2n B.3+(n+1)×2nC.1+(n+1)×2n D.1+(n-1)×2n8.[2024·商丘二模]已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿意a1=1,an+1-an≥2(n∈N*),則 ()A.an≥2n+1 B.Sn≥n2C.an≥2n-1 D.Sn≥2n-19.[2024·廣東珠海一中等六校聯(lián)考]數(shù)列{an}滿意a1=1,且對于隨意的n∈N*,都有an+1=an+a1+n,則1a1+1a2+…+1A.20162017 B.40322017 C.20172018 10.已知數(shù)列{an}滿意a1=43,an+1-1=an2-an,則m=1a1+1a2+A.1 B.2C.3 D.411.[2024·大連一模]已知數(shù)列{an}滿意an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=.

12.[2024·成都三診]在遞減的等差數(shù)列{an}中,a1a3=a22-4,a1=13,則數(shù)列1ana13.[2024·宜賓二診]數(shù)列{an}的通項(xiàng)公式為an=ncos2nπ4-sin2nπ4,14.[2024·貴陽模擬]已知Sn為數(shù)列{an}的前n項(xiàng)和,且a1=3,Sn=an+n2-1(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)設(shè)bn=1anan+1,求數(shù)列{bn}的前15.[2024·內(nèi)江一模]設(shè)數(shù)列{an}滿意a1+2a2+4a3+…+2n-1an=n.(1)求數(shù)列{an}的通項(xiàng)公式;(2)求數(shù)列{an+log2an}的前n項(xiàng)和Sn.16.[2024·南昌三模]已知數(shù)列{an}滿意a12+a222+a32(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=(-1)nan2,求數(shù)列{bn}課時(shí)作業(yè)(二十九)1.B[解析]由S8=4S4,得8a1+28d=4(4a1+6d),解得a1=12,所以a10=a1+9=192.D[解析]由Sn=-n2-n可知,當(dāng)n≥2時(shí),an=Sn-Sn-1=-n2-n-[-(n-1)2-(n-1)]=-2n,當(dāng)n=1時(shí),a1=S1=-2,上式也成立,所以an=-2n,所以2(n+1)an=-22n(n+1)=-1n-1n+1,所以數(shù)列2(n+1)an的前n項(xiàng)和Tn=-1-12+12-13+…+1n-3.D[解析]由n+1an+1=nan+2n,得n則nan-n-1an-1=2n-1,n-1an-1-n將以上各式相加得nan-1a1=21+22+…+2n-1=2n-2,又a1=12,所以a因此S100=1×12+2×122+…+100則12S100=1×122+2×123+…+99×1兩式相減得12S100=12+122+…+12所以S100=2-1299-100×12100=2-51299.故選D.4.B[解析]設(shè)等差數(shù)列{an}的公差為d,∵a9=12a12+6,a2=4,∴解得a1=d=2,∴Sn=2n+n(n-1)2∴1Sn=1n(n∴1S1+1S2+…+1S10=1-12+12-13+…+110-111=1故選B.5.A[解析]由題意得a1+a2+…+a2024=(a1+a2)+(a3+a4)+…+(a2024+a2024)=(1-4)+(7-10)+…+[(3×2024-2)-(3×2024-2)]=(-3)×1009=-3027.6.C[解析]因?yàn)閍n+1=Sn+1-Sn,所以Sn+1=3Sn,又S1=a1=1,故數(shù)列{Sn}是首項(xiàng)為1,公比為3的等比數(shù)列,則Sn=S1·3n-1=3n-1,故選C.7.D[解析]當(dāng)q=1時(shí),不符合題意.當(dāng)q≠1時(shí),a1(1-q3)1-q=7,a1(1-q6)1-q=63,解得設(shè)數(shù)列{nan}的前n項(xiàng)和為Tn,則Tn=1+2×2+3×22+…+n·2n-1,2Tn=1×2+2×22+…+(n-1)·2n-1+n·2n,兩式相減得-Tn=1+2+22+…+2n-1-n·2n=1-2n1-2-n·2n=(1-n)·2n-1,所以Tn=1+(n-1)·8.B[解析]由題得a2-a1≥2,a3-a2≥2,a4-a3≥2,…,an-an-1≥2(n≥2),∴a2-a1+a3-a2+a4-a3+…+an-an-1≥2(n-1)(n≥2),∴an-a1≥2(n-1)(n≥2),∴an≥2n-1(n≥2),∴a2≥3,a3≥5,…,an≥2n-1,又a1=1,∴a1+a2+a3+…+an≥1+3+5+…+(2n-1),∴Sn≥n2(1+2n-1)=n2,故選B9.D[解析]由題意可得,an+1-an=n+1,則a2-a1=2,a3-a2=3,…,an-an-1=n(n≥2),以上各式相加可得an=n(n+1)2(n≥2),又a1=1,符合上式,所以an=n(n+1)2,所以1a1+1a2+…+1a2017=2×1-12+12-13+…+12017-12018故選D.10.B[解析]∵a1=43,an+1-1=an2-an,∴an>1,∴an+1-an=(an-1)2>0,∴an+1>an,∴數(shù)列{an}是遞增數(shù)列.由an+1-1=an2-an=an(an-1),得1an+1-1=1an∴m=1a1+1a2+…+1a2017=1a1-1-1a2-1+1a2-1-1a3-1+…+1a2017-1-1a2018-1=1a1-1-1a2018-1=3-1a2018-1.∵a1=43>1,且數(shù)列{11.18[解析]由題意得,數(shù)列{an}為等差數(shù)列,且an=-5+2(n-1)=2n-7,因此|a1|+|a2|+…+|a6|=5+3+1+1+3+5=18.12.613[解析]設(shè)公差為d,則d<0.由a1a3=a22-4,a1=13,得13(13+2d)=(13+d)2-4,得d=-2,則an=13-2(n-1)=15所以1anan+1=1(15-2n)(13-2n)=1212n-15-12n-13,所以數(shù)列1a13.20[解析]由題意得,an=ncos2nπ4-sin2nπ4=n·cos則a1=0,a2=-2,a3=0,a4=4,a5=0,a6=-6,a7=0,a8=8,…,所以a1+a2+a3+a4=2,a5+a6+a7+a8=2,…,所以S40=10×2=20.14.解:(1)由Sn=an+n2-1①,得Sn+1=an+1+(n+1)2-1②.②-①得an+1=Sn+1-Sn=an+1-an+(n+1)2-n2,整理得an=2n+1.(2)由an=2n+1可知bn=1(2n+1)(2n+3)則Tn=b1+b2+…+bn=12×13-15+15-17+…+12n+1-12n+315.解:(1)∵數(shù)列{an}滿意a1+2a2+4a3+…+2n-1an=n,∴當(dāng)n≥2時(shí),a1+2a2+4a3+…+2n-2an-1=n-1,∴當(dāng)n≥2時(shí),2n-1an=1,即an=12n-1,當(dāng)n=1時(shí),a1=1也滿意上式,∴數(shù)列{an}的通項(xiàng)公式為an=12n-(2)由(1)知,an+log2an=12n-1∴Sn=(a1+log2a1)+(a2+log2a2)+(a3+log2a3)+…+(an+log2an)=(1-0)+12-1+122-2+…+12n-1+1-n=1+12+122+…+12n-1-[1+2+3+…+(n-1)]=2-16.解:(1)∵a12+a222+a323+∴當(dāng)n≥2時(shí),a12+a222+a323+…+a①-②得an2n=2n(n≥2),∴an=n·2n+1(n又∵當(dāng)n=1時(shí),a12