新教材2024-2025學(xué)年高中數(shù)學(xué)第五章數(shù)列5.3等比數(shù)列5.3.2等比數(shù)列的前n項(xiàng)和分層作業(yè)新人教B版選擇性必修第三冊(cè)

第五章5.3.2等比數(shù)列的前n項(xiàng)和A級(jí)必備學(xué)問(wèn)基礎(chǔ)練1.[探究點(diǎn)一]已知等比數(shù)列{an}各項(xiàng)均為正數(shù),a3,a5,-a4成等差數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和,則S6S3=A.2 B.78 C.98 D2.[探究點(diǎn)一]已知等比數(shù)列{an}中,a1+a4=2,a2+a5=4,則數(shù)列{an}的前6項(xiàng)和S6的值為()A.12 B.14 C.16 D.183.[探究點(diǎn)二]在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,若am+1am-1=2am(m≥2),數(shù)列{an}的前n項(xiàng)積為Tn,若T2m-1=512,則m的值為()A.4 B.5 C.6 D.74.[探究點(diǎn)一]古希臘哲學(xué)家芝諾提出了一個(gè)悖論:讓阿基里斯和烏龜賽跑,他的速度是烏龜速度的10倍,烏龜在他前面100米爬行,他在后面追,但他不行能追上烏龜.緣由是在競(jìng)賽中,追者首先必需到達(dá)被追者的動(dòng)身點(diǎn),當(dāng)阿基里斯追了100米時(shí),烏龜已在他前面爬行了10米,而當(dāng)他追到烏龜爬行的10米處時(shí),烏龜又向前爬行了1米,就這樣,烏龜總能領(lǐng)先一段距離,不管這個(gè)距離有多小,只要烏龜不停地向前爬行,阿基里斯就恒久追不上烏龜.試問(wèn)在阿基里斯與烏龜?shù)母?jìng)賽中,當(dāng)阿基里斯與烏龜相距0.01米時(shí),烏龜共爬行了()A.11.1米 B.10.1米 C.11.11米 D.11米5.[探究點(diǎn)一·2024陜西銅川校考一模]設(shè)各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和為Sn,若S3=13,S6=364,則通項(xiàng)an為()A.32n-1 B.32n C.3n D.3n-16.[探究點(diǎn)一、二](多選題)[2024湖北武漢洪山高級(jí)中學(xué)高二階段練習(xí)]已知Sn為數(shù)列{an}的前n項(xiàng)和,下列說(shuō)法肯定正確的是()A.若{an}為等差數(shù)列,則S5,S10-S5,S15-S10為等差數(shù)列B.若{an}為等比數(shù)列,則S5,S10-S5,S15-S10為等比數(shù)列C.若{an}為等差數(shù)列,則S5D.若{an}為等比數(shù)列,則S57.[探究點(diǎn)一]已知等比數(shù)列{an}的公比為2,前n項(xiàng)和為Sn,若S5=1,則S10=.

8.[探究點(diǎn)一]在等比數(shù)列{an}中,若an=2n-1,則a12+a22+9.[探究點(diǎn)三·2024江蘇南京高二期末]求和:Sn=1+(1+12)+(1+12+14)+(1+12+14+18)+…+10.[探究點(diǎn)一]已知數(shù)列{an}是首項(xiàng)為1,公差為2的等差數(shù)列,Sn為數(shù)列{an}的前n項(xiàng)和.(1)求an及Sn;(2)設(shè)數(shù)列{bn}是首項(xiàng)為2的等比數(shù)列,其公比q滿意q2-(a4+1)q+S4=0,求數(shù)列{bn}的通項(xiàng)公式及其前n項(xiàng)和Tn.11.[探究點(diǎn)三·2024湖南湘潭高三期末]已知等差數(shù)列{an}和等比數(shù)列{bn}滿意a1=2,b1=1,a2+a3=10,b2b3=-a4.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)設(shè)cn=an+bn,求c1+c3+c5+…+c2n-1.B級(jí)關(guān)鍵實(shí)力提升練12.設(shè)數(shù)列{an}的各項(xiàng)均為正數(shù),前n項(xiàng)和為Sn,log2an+1=1+log2an,且a3=4,則S6的值為()A.128 B.65 C.64 D.6313.記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a5-a3=12,a6-a4=24,則SnanA.2n-1 B.2-21-nC.2-2n-1 D.21-n-114.等比數(shù)列{an}的項(xiàng)數(shù)為奇數(shù),全部奇數(shù)項(xiàng)的和S奇=255,全部偶數(shù)項(xiàng)的和S偶=-126,末項(xiàng)是192,則首項(xiàng)a1的值為()A.1 B.2 C.3 D.415.[北師大版教材習(xí)題改編]設(shè)數(shù)列{an}是由正數(shù)組成的等比數(shù)列,公比q=2,且a1a2a3…a30=230,那么a3a6…a30=()A.210 B.215 C.220 D.21616.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,若S10S5=1A.13 B.12 C.2317.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,若S4=3,S8=9,則S16的值為.

18.已知數(shù)列{an}是以1為首項(xiàng),2為公差的等差數(shù)列,{bn}是以1為首項(xiàng),2為公比的等比數(shù)列,設(shè)cn=abn,Tn=c1+c2+…+cn(n∈N+),則當(dāng)Tn<2023時(shí),n的最大值為19.設(shè)等比數(shù)列{an}滿意a1+a2=4,a3-a1=8.(1)求數(shù)列{an}的通項(xiàng)公式;(2)記Sn為數(shù)列{log3an}的前n項(xiàng)和.若Sm+Sm+1=Sm+3,求m.C級(jí)學(xué)科素養(yǎng)創(chuàng)新練20.已知等比數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,Sn+1+2Sn-1=3Sn(n≥2).(1)求數(shù)列{an}的通項(xiàng)公式;(2)令bn=an+1SnSn+1,求數(shù)列{b

5.3.2等比數(shù)列的前n項(xiàng)和1.C設(shè)等比數(shù)列{an}的公比為q,則有q>0,又a3,a5,-a4成等差數(shù)列,∴a3-a4=2a5,∴a1q2-a1q3=2a1q4,即1-q=2q2,解得q=-1(舍去)或q=12,∴q=1∴S6S3=a1(1-2.B設(shè)數(shù)列{an}的公比為q,則q=a2+∴a1+a4=a1+a1q3=9a1=2,∴a1=29,∴S6=29×故選B.3.B因?yàn)閿?shù)列{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,所以am+1am-1=2am=am2,則am=2.又T2m-1=a1a2…a2m-1=am2m-1,所以22m-1=512=24.C依題意,烏龜爬行的距離組成等比數(shù)列{an},其首項(xiàng)a1=10,公比q=0.1,前n項(xiàng)和為Sn,所以當(dāng)an=0.01時(shí),Sn=a1-anq故選C.5.D設(shè){an}的公比為q,由題可知q>0,且q≠1.由題可得S3=a1(1-q3)1-q=13,故選D.6.ABCA選項(xiàng),{an}為等差數(shù)列,設(shè)公差為d,所以S5=5a1+10d,S10=10a1+45d,S15=15a1+105d,故S10-S5=5a1+35d,S15-S10=5a1+60d,因?yàn)?(S10-S5)=S5+S15-S10,所以S5,S10-S5,S15-S10成等差數(shù)列,A正確;B選項(xiàng),{an}成等比數(shù)列,設(shè)公比為q,若q=1,則S5=5a1,S10=10a1,S15=15a1,則S10-S5=5a1,S15-S10=5a1,故(S10-S5)2=S5(S15-S10),故S5,S10-S5,S15-S10成等比數(shù)列.若q≠1,則S5=a1(1-q5)1-q,所以S10-S5=a1S15-S10=a1則S10-S5S5=故S10-S5S5=S15-S10S10-綜上,若{an}為等比數(shù)列,則S5,S10-S5,S15-S10肯定為等比數(shù)列,B正確;C選項(xiàng),{an}為等差數(shù)列,設(shè)公差為d,則S55=5a1+10d5=a1+2d,S1010=10因?yàn)镾1010-S5故S10則S5D選項(xiàng),{an}成等比數(shù)列,設(shè)公比為q.若q=1,則S55=a1,S1010=a1,則S55若q≠1,則S5則S10因?yàn)?1-q20)(1-q5)=1-q5-q20+q25≠(1所以S55綜上,若{an}為等比數(shù)列,則S55故選ABC.7.33∵S5=a1(1-25)1-2=31∴S10=a1(1-28.13(4n-1)∵an=2n-1,∴an2=4n-1∴a12+a22+…+9.2n+12n-1-2設(shè)數(shù)列{an}的通項(xiàng)公式為an=1+12+14+…+12n-1=1-(所以Sn=21-12+1-122+…+1-12n=2n-(12+122+123+…+12n)=2n-10.解(1)設(shè){an}的公差為d,由題可知an=a1+(n-1)d=2n-1,Sn=n(a1(2)由(1)得a4=7,S4=16.因?yàn)閝2-(a4+1)q+S4=0,即q2-8q+16=0,所以(q-4)2=0,從而q=4,所以bn=b1qn-1=2·4n-1=22n-1,Tn=b1(1-q11.解(1)設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q,則a2+a3=a1+d+a1+2d=4+3d=10,解得d=2,∴an=a1+(n-1)d=2+2(n-1)=2n,∴b2b3=b1qb1q2=q3=-a4=-8,解得q=-2,∴bn=b1qn-1=(-2)n-1,即an=2n,bn=(-2)n-1.(2)由(1)知{an}為等差數(shù)列,{bn}為公比q=-2的等比數(shù)列,∴a1,a3,a5,…,a2n-1為等差數(shù)列,b1,b3,b5,…,b2n-1為公比為q2的等比數(shù)列.∵cn=an+bn,∴c1+c3+c5+…+c2n-1=(a1+a3+a5+…+a2n-1)+(b1+b3+b5+…+b2n-1)=n(a1+a2n12.D因?yàn)閘og2an+1=1+log2an,所以log2an+1=log22an,所以an+1=2an,所以數(shù)列{an}是公比為2的等比數(shù)列.又因?yàn)閍3=4,所以a1=a32S6=1×(113.B設(shè)等比數(shù)列{an}的公比為q.∵a5-a3=12,a6-a4=24,∴a6-a又a5-a3=a1q4-a1q2=12a1=12,∴a1=1.∴an=a1qn-1=2n-1,Sn=a1(1-q∴Snan=2n-12n-故選B.14.C設(shè)等比數(shù)列{an}共有2k+1(k∈N+)項(xiàng),公比為q,則a2k+1=192,則S奇=a1+a3+…+a2k-1+a2k+1=1q(a2+a4+…+a2k)+a2k+1=1qS偶+a2k+1=-126q+192=255,解得q=-2,而S奇=a1-a2k+115.C因?yàn)閍1a2a3…a30=a130q1+2+…+29=(a110q145)3,a3a6…a30=a110q2+5+…+所以a3a6…a30=(a1a2a3…a30)13q因?yàn)閍1a2a3…a30=230,q=2,所以a3a6…a30=220.16.D設(shè)等比數(shù)列{an}的公比為q,q≠0.若q=1,則an=a1,所以Sn=na1,所以S10S所以q≠1,Sn=a1所以S10S5=a1(1-q10)1-故選D.17.45設(shè)等比數(shù)列{an}的公比為q.因?yàn)镾4≠0,所以由等比數(shù)列前n項(xiàng)和的性質(zhì)可知,S4,S8-S4,S12-S8,S16-S12成等比數(shù)列,且公比為S8-所以S12-S8=3×22=12,S16-S12=3×23=24,因此S16=S4+(S8-S4)+(S12-S8)+(S16-S12)=3+6+12+24=45.18.9∵數(shù)列{an}是以1為首項(xiàng),2為公差的等差數(shù)列,∴an=2n-1.∵數(shù)列{bn}是以1為首項(xiàng),2為公比的等比數(shù)列,∴bn=2n-1.∴Tn=c1+c2+…+cn=ab1+ab2+…+abn=a1+a2+a4+…+a2n-1=(2×1-1)+(2×2-1)+(2×4-1)+…+(2×2n-1-1)=2(1+2+4+…+2n-1∵Tn<2024,∴2n+1-n-2<2024,∴n≤9.故當(dāng)Tn<2024時(shí),n的最大值是9.19.解(1)設(shè){an}的公比為q,則an=a1qn-1.由已知得a1+所以{an}的通項(xiàng)公式為an=3n-1.(2