适用于新高考新教材广西专版2025届高考数学一轮总复习第六章数列课时规范练32数列求和

课时规范练32数列求和基础巩固组1.(2023广东江门一模)已知等差数列{an}的前n项和为Sn,公差d<0,a10a9<-1,则使得Sn>0的最大整数n为A.9 B.10 C.17 D.182.Sn=12+222+38A.2n+1-C.2n-n3.已知Sn是数列{an}的前n项和,a1=1,a2=2,a3=3,记bn=an+an+1+an+2且bn+1-bn=2,则S31=()A.171 B.278C.351 D.3954.在数列{an}中,an+an+1=2n,Sn为其前n项和,若a1=a4,则S101=()A.4882 B.5100C.5102 D.52125.(2023四川攀枝花二模)已知正项数列{an}的前n项和为Sn,且2anSn=1+an2,设bn=log2Sn+1Sn,数列{bn}的前n项和为Tn,则满足Tn≥2A.15 B.16 C.3 D.46.已知数列{an}满足an=1+2+4+…+2n-1,则数列2nanan+1的前57.①{2nan}为等差数列,且a3=58;②an2n-1为等比数列,且a2=34.从①②两个条件中任选一个在数列{an}中,a1=12,.(1)求{an}的通项公式;(2)已知{an}的前n项和为Sn,试问是否存在正整数p,q,r,使得Sn=p-qan+r?若存在,求p,q,r的值;若不存在,请说明理由.8.已知数列{an}的前n项和Sn满足Sn=3n2-n2,数列{log3bn}是公差为-1的等差数列,(1)求数列{an},{bn}的通项公式;(2)设cn=a2n+1+b2n+1,求数列{cn}的前n项和Tn.综合提升组9.已知数列{an}的前n项和为Sn=2-3n,则此数列奇数项的前m项和为()A.94-9C.94-9m10.对于实数x,[x]表示不超过x的最大整数.已知数列{an}的通项公式为an=1n+1+n,前n项和为Sn,则[S1]+[S2]+…+[S40]A.105 B.120 C.125 D.13011.在数列{an}中,a1=1,nan+1=(n+1)an+n(n+1),若bn=ancos2nπ3,且数列{bn}的前n项和为Sn,则S11=A.64 B.80 C.-64 D.-8012.在数列{an}中,a1=2,ap+q=apaq(p,q∈N*),记bm为数列{an}中在区间(0,m](m∈N*)内的项的个数,则数列{bm}的前150项和S150=.

13.(2023山西联考模拟)已知数列{an}满足2an-3an+1+an+2=n-1.(1){an+1-an+n}是否为等比数列?并说明理由;(2)若a1=a2=1,求{an}的通项公式.14.在等差数列{an}中,a1+3,a3,a4成等差数列,且a1,a3,a8成等比数列.(1)求数列{an}的通项公式;(2)在任意相邻两项ak与ak+1(k=1,2,…)之间插入2k个2,使它们和原数列的项构成一个新的数列{bn},求数列{bn}的前200项和T200.创新应用组15.在数列{an}中,a1=3,an+1=3an-4n,若bn=4n2+8n+5anan+1,且数列{bn}的前n项和为A.n1+22n+3 B.C.n1+16n+9 D.n1+2616.(2023吉林三模)已知数列{an}满足an=2n-2,n为奇数,3n-2,n为偶数,{a(1)求a1,a2,并判断1024是数列中的第几项;(2)求S2n-1.17.已知等比数列{an}的前n项和为Sn,公比q>0,S2=2a2-2,S3=a4-2,数列{an}满足a2=4b1,nbn+1-(n+1)bn=n2+n(n∈N*).(1)求数列{an}的通项公式;(2)证明:数列bnn为等差数列;(3)设数列{cn}的通项公式为cn=-anbn2,n为奇数,anb

课时规范练32数列求和1.C解析∵a10a9<-1,∴a10,a∵d<0,∴a9>0,a10<0,∴a10<-a9,即a10+a9<0.∵S17=17(a1+a17)2=17a9>0,S18=18(a1+a18)2=9(a9+a102.A解析由Sn=12+222+323+…+n2n,可得12Sn=122+223+…+n-12n3.C解析由bn+1-bn=2,得an+1+an+2+an+3-(an+an+1+an+2)=an+3-an=2,∴a1,a4,a7,…是首项为1,公差为2的等差数列,a2,a5,a8,…是首项为2,公差为2的等差数列,a3,a6,a9,…是首项为3,公差为2的等差数列,S31=(a1+a4+…+a31)+(a2+a5+…+a29)+(a3+a6+…+a30)=1×11+11×10×22+2×10+10×9×24.C解析因为an+an+1=2n,①所以an+1+an+2=2n+2,②由②-①得an+2-an=2,所以数列{an}奇数项与偶数项均成公差为2的等差数列.当n为奇数时,an=a1+n-12×2=n+a当n为偶数时,an=a2+n-22×2=n+a2-2=n+(2-a1)-2=n-a1.又因为a1=a4,所以a1=4-a1,得a1=2,所以a所以S101=(a1+…+a101)+(a2+…+a100)=512×(2+102)+502×(0+98)=5102.故选5.A解析由题意知2Sn=1an+an且an>0,当n=1时,2S1=1a1+a1,则S1当n=2时,2S2=2(a1+a2)=1a2+a2,则a22+2a2-1=0,可得a2=2-1,∴S2=a1+a当n≥2时,将an=Sn-Sn-1代入2anSn=1+an2,可得2(Sn-Sn-1)Sn=1+(Sn-Sn-1)2,即2Sn2-2SnSn-1=1+Sn2-2SnSn-1+又S22-S12=1也成立,故{Sn2}是首项、公差均为1的等差数列,则S又Tn=b1+b2+…+bn=log2S2S1+log2S3S2+…+log2Sn+1Sn=log2Sn+1S1=所以n+1≥16,即n≥15,故Tn≥2的n的最小正整数解为15.故选A.6.6263解析因为an=1+2+4+…+2n-1=2n-1,an+1=2n+1-1,所以2n所以2nanan+1的前5项和为121-1-122-1+122-7.解(1)若选①:设等差数列{2nan}的公差为d,则d=23a3-2a13-1=5-12=2,∴2nan=2a若选②:设等比数列an2n-1则q=a22×2-1a12×1-1=12,∴an2(2)Sn=12+32212Sn=122+3则两式相减得12Sn=12+2×122+123+∴Sn=3-2n∵Sn=3-2n+32n=3-4×2(n+2)-∴存在正整数p,q,r,使得Sn=p-qan+r,且p=3,q=4,r=2.8.解(1)当n=1时,a1=S1=3-12=1,当n≥2,an=Sn-Sn-1=3n2-n2-3(n-1)2-(n-1)2=3n-2,所以n=1满足n≥2时的情况,所以an=3n-2(n∈N*).因为log3bn=log3(2)因为cn=a2n+1+b2n+1=3(2n+1)-2+31-(2n+1)=6n+1+19n,所以Tn=(7+6n+1)n2+19[1-(19)

n]9.B解析当n≥2时,an=Sn-Sn-1=(2-3n)-(2-3n-1)=-2·3n-1.因为当n=1时,a1=-1不满足,所以数列{an}从第2项开始成等比数列.又a3=-18,则数列{an}的奇数项构成的数列的前m项和Tm=-18(1-9m-110.B解析因为an=1n所以Sn=2-1+3-2+[S1]=[1+1-1]=0,[S2]=[2+1-1]=0,[S3]=[3+1-1]=1,[S4]=[4+1-1]=1,…,[S7]=[7+1-1]=1,[S8]=[8+1-1]=2,[S9]=[9+1-1]=2,…,[S14]=[14+1-1]=2,[S15]=[15+1-1]=3,[S16]=[16+1-1]=3,…,[S23]=[23+1-1]=3,[S24]=[24+1-1]=4,[S25]=[25+1-1]=4,…,[S34]=[34+1-1]=4,[S35]=[35+1-1]=5,[S36]=[36+1-1]=5,…,[S40]=[40+1-1]=5.故[S1]+[S2]+…+[S40]=0×2+1×5+2×7+3×9+4×11+5×6=120.故选B.11.C解析已知nan+1=(n+1)an+n(n+1),则an+1n+1=ann+1,可得数列ann是首项为1,公差为1的等差数列,即有ann=n,即an=n2,则b则S11=-12(12+22+42+52+72+82+102+112)+(32+62+92)=-12(12+22-32-32+42+52-62-62+72+82-92-92+102+112)=-12×(5+23+41+59)=-64.12.803解析令p=1,q=n,则a1+n=a1an=2an,所以数列{an}是首项为2,公比为2的等比数列,所以an=2n.当m=1时,b1=0.当2n≤m<2n+1时,bm=n,所以S150=b1+(b2+b3)+(b4+b5+b6+b7)+…+(b64+b65+…+b127)+(b128+b129+…+b150)=0+1×2+2×22+3×23+4×24+5×25+6×26+7×(150-127)=803.13.解(1)由2an-3an+1+an+2=n-1,得an+2-an+1+(n+1)=2an+1-2an+2n=2(an+1-an+n),当a2-a1+1=0时,数列{an+1-an+n}为各项为0的常数列,不是等比数列;当a2-a1+1≠0时,数列{an+1-an+n}是以a2-a1+1为首项,2为公比的等比数列.(2)当a1=a2=1时,a2-a1+1=1,由(1)知,数列{an+1-an+n}是以1为首项,2为公比的等比数列,∴an+1-an+n=2n-1,即an+1-an=2n-1-n,当n≥2时,an=(an-an-1)+(an-1-an-2)+(an-2-an-3)+…+(a3-a2)+(a2-a1)+a1=2n-2-(n-1)+2n-3-(n-2)+2n-4-(n-3)+…+21-2+20-1+1=(20+21+…+2n-2)-[1+2+…+(n-1)]+1=1-2n-11-2-又a1=1满足an=2n-1-n(∴an=2n-1-n(n-1)2(14.解(1)设等差数列{an}的公差为d.由题意得a1+3+a4=2a3,即2a1+3+3d=2a1+4d,解得d=3.又a1a8=a32,即a1(a1+7×3)=(a1+2×3)2,解得a1=4,所以an=3n+(2)在数列{bn}中,ak+1前面(包括ak+1)共有2+22+23+…+2k+(k+1)=2k+1+k-1项.令2k+1+k-1≤200(k=1,2,…),则k≤6,所以a1,a2,a3,a4,a5,a6,a7出现在数列{bn}的前200项中.当k=6时,2k+1+k-1=133,所以a7前面(包括a7)共有133项,所以a7后面(不包括a7)还有67个2.所以T200=(4+7+…+22)+2(2+22+23+…+26+67)=91+386=477.15.D解析由an+1=3an-4n,可得an+1-[2(n+1)+1]=3[an-(2n+1)].∵a1=3,∴a1-(2×1+1)=0,则可得数列{an-(2n+1)}为常数列0,即an-(2n+1)=0,∴an=2n+1,∴bn=4n2+8n+5(2n+1∴Sn=n+13-15+15-17+…+12n+1-16.解(1)由an=2n-2,n为奇数,3n-2,令2n-2=1024=210,解得n=12为偶数,不符合题意,舍去;令3n-2=1024,解得n=342,符合题意.因此1024是数列{an}的第342项.(2)由题意得a2n-1=22n-3,又a2n所以数列{a2n-1}是以12为首项,4为公比的等比数列a2n=6n-2,又a2n+2-a2n=6,所以数列{a2n}是以4为首项,6为公差的等差数列.S2n-1为数列{a2n-1}的前n项和与数列{a2n}的前n-1项和的总和.故S2n-1=12(1-4n)1-4+(n-1)(4+6n-8)17.(1)解因为等比数列{an}的前n项和为Sn,公比q>0,S2=2a2-2,S3=a4-2,所以S3-S2=a4-2a2=a3,整理得a2q2-2a2=a2q.又a2≠0,所以q2-q-2=0.由于q>0,解得q=2.由于a1+a2=2a2-2,解得a1=2,所以an=2n.(2)证明数列{an}满足a2=4b1,解得b1=1.由于nbn+1-(n+1)bn=n2+n,所以