2020版高考数学复习第五单元第29讲数列求和练习文（含解析）新人教A版.docx

第29讲数列求和 1.2018南宁模拟 已知an是公差为1的等差数列,Sn为数列an的前n项和,若S8=4S4,则a10=()A.172B.192C.10D.122.2018潍坊二模 设数列an的前n项和为Sn,若Sn=-n2-n,则数列2(n+1)an的前40项的和为()A.3940B.-3940C.4041D.-40413.2018山西榆社中学模拟 设Sn为数列an的前n项和,已知a1=12,n+1an+1=nan+2n,则S100=()A.2-492100B.2-49299C.2-512100D.2-512994.2018成都第七中学一诊 已知等差数列an的前n项和为Sn,a9=12a12+6,a2=4,则数列1Sn的前10项和为()A.1112B.1011C.910D.895.2018江淮十校三联 若数列an的通项公式是an=(-1)n+1(3n-2),则a1+a2+a2018=()A.-3027B.3027C.-3030D.30306.2018株洲一模 已知数列an的前n项和为Sn,且a1=1,2Sn=an+1,则Sn=()A.2n-1B.2n-1C.3n-1D.12(3n-1)7.已知等比数列an的前n项和为Sn,若S3=7,S6=63,则数列nan的前n项和为()A.-3+(n+1)2nB.3+(n+1)2nC.1+(n+1)2nD.1+(n-1)2n8.2018商丘二模 已知数列an的前n项和为Sn,且满足a1=1,an+1-an2(nN*),则()A.an2n+1B.Snn2C.an2n-1D.Sn2n-19.2018广东珠海一中等六校联考 数列an满足a1=1,且对于任意的nN*,都有an+1=an+a1+n,则1a1+1a2+1a2017等于()A.20162017B.40322017C.20172018D.4034201810.已知数列an满足a1=43,an+1-1=an2-an,则m=1a1+1a2+1a2017的整数部分是()A.1B.2C.3D.411.2018大连一模 已知数列an满足an+1-an=2,a1=-5,则|a1|+|a2|+|a6|=.12.2018成都三诊 在递减的等差数列an中,a1a3=a22-4,a1=13,则数列1anan+1的前n项和的最大值为.13.2018宜宾二诊 数列an的通项公式为an=ncos2n4-sin2n4,其前n项和为Sn,则S40=.14.2018贵阳模拟 已知Sn为数列an的前n项和,且a1=3,Sn=an+n2-1(nN*).(1)求数列an的通项公式;(2)设bn=1anan+1,求数列bn的前n项和Tn.15.2018内江一模 设数列an满足a1+2a2+4a3+2n-1an=n.(1)求数列an的通项公式;(2)求数列an+log2an的前n项和Sn.16.2017南昌三模 已知数列an满足a12+a222+a323+an2n=n2+n.(1)求数列an的通项公式;(2)若bn=(-1)nan2,求数列bn的前n项和Sn.课时作业(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=192.2.D解析 由Sn=-n2-n可知,当n2时,an=Sn-Sn-1=-n2-n-(n-1)2-(n-1)=-2n,当n=1时,a1=S1=-2,上式也成立,所以an=-2n,所以2(n+1)an=-22n(n+1)=-1n-1n+1,所以数列2(n+1)an的前n项和Tn=-1-12+12-13+1n-1n+1=-1-1n+1=-nn+1,所以其前40项和为T40=-4041,故选D.3.D解析 由n+1an+1=nan+2n,得n+1an+1-nan=2n,则nan-n-1an-1=2n-1,n-1an-1-n-2an-2=2n-2,2a2-1a1=21,将以上各式相加得nan-1a1=21+22+2n-1=2n-2,又a1=12,所以an=n12n,因此S100=112+2122+10012100,则12S100=1122+2123+9912100+10012101,两式相减得12S100=12+122+12100-10012101,所以S100=2-1299-10012100=2-51299.故选D.4.B解析 设等差数列an的公差为d,a9=12a12+6,a2=4,a1+8d=12(a1+11d)+6,a1+d=4,解得a1=d=2,Sn=2n+n(n-1)22=n2+n,1Sn=1n(n+1)=1n-1n+1,1S1+1S2+1S10=1-12+12-13+110-111=1-111=1011.故选B.5.A解析 由题意得a1+a2+a2018=(a1+a2)+(a3+a4)+(a2017+a2018)=(1-4)+(7-10)+(32017-2)-(32018-2)=(-3)1009=-3027.6.C解析 因为an+1=Sn+1-Sn,所以Sn+1=3Sn,又S1=a1=1,故数列Sn是首项为1,公比为3的等比数列,则Sn=S13n-1=3n-1,故选C.7.D解析 当q=1时,不符合题意.当q1时,a1(1-q3)1-q=7,a1(1-q6)1-q=63,解得q=2,a1=1,所以an=a1qn-1=2n-1,所以nan=n2n-1.设数列nan的前n项和为Tn,则Tn=1+22+322+n2n-1,2Tn=12+222+(n-1)2n-1+n2n,两式相减得-Tn=1+2+22+2n-1-n2n=1-2n1-2-n2n=(1-n)2n-1,所以Tn=1+(n-1)2n,故选D.8.B解析 由题得a2-a12,a3-a22,a4-a32,an-an-12(n2),a2-a1+a3-a2+a4-a3+an-an-12(n-1)(n2),an-a12(n-1)(n2),an2n-1(n2),a23,a35,an2n-1,又a1=1,a1+a2+a3+an1+3+5+(2n-1),Snn2(1+2n-1)=n2,故选B.9.D解析 由题意可得,an+1-an=n+1,则a2-a1=2,a3-a2=3,an-an-1=n(n2),以上各式相加可得an=n(n+1)2(n2),又a1=1,符合上式,所以an=n(n+1)2,则1an=21n-1n+1,所以1a1+1a2+1a2017=21-12+12-13+12017-12018=40342018.故选D.10.B解析a1=43,an+1-1=an2-an,an1,an+1-an=(an-1)20,an+1an,数列an是递增数列.由an+1-1=an2-an=an(an-1),得1an+1-1=1an(an-1)=1an-1-1an,1an=1an-1-1an+1-1,m=1a1+1a2+1a2017=1a1-1-1a2-1+1a2-1-1a3-1+1a2017-1-1a2018-1=1a1-1-1a2018-1=3-1a2018-1.a1=431,且数列an是递增数列,a2=1+49,a3=1+5281,a4=1+691665612,a20182,01a2018-11,2m3,m的整数部分是2,故选B.11.18解析 由题意得,数列an为等差数列,且an=-5+2(n-1)=2n-7,因此|a1|+|a2|+|a6|=5+3+1+1+3+5=18.12.613解析 设公差为d,则d0.由a1a3=a22-4,a1=13,得13(13+2d)=(13+d)2-4,得d=-2,则an=13-2(n-1)=15-2n.所以1anan+1=1(15-2n)(13-2n)=1212n-15-12n-13,所以数列1anan+1的前n项和Sn=12-113-12n-13,则当n=6时,Sn取得最大值613.13.20解析 由题意得,an=ncos2n4-sin2n4=ncosn2,则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=102=20.14.解:(1)由Sn=an+n2-1,得Sn+1=an+1+(n+1)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)=1212n+1-12n+3,则Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).15.解:(1)数列an满足a1+2a2+4a3+2n-1an=n,当n2时,a1+2a2+4a3+2n-2an-1=n-1,当n2时,2n-1an=1,即an=12n-1,当n=1时,a1=1也满足上式,数列an的通项公式为an=12n-1(nN*).(2)由(1)知,an+log2an=12n-1+1-n,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)=2-12n-1-n22+n2.16.解:(1)a12+a222+a323+an2n=n2+n,当n2时,a12+a222+a323+an-12n-1=(n-1)2+n-1,-得an2n=2n(n2),an=n2n+1(n2).又当n=1时,a