2023屆高考數(shù)學(xué)一輪復(fù)習(xí)第六章數(shù)列考點(diǎn)規(guī)范練30等比數(shù)列及其前n項(xiàng)和文新人教A版

考點(diǎn)標(biāo)準(zhǔn)練30等比數(shù)列及其前n項(xiàng)和根底穩(wěn)固1.等比數(shù)列{an}滿足a1=,a3a5=4(a4-1),那么a2=A.2 B.1 C. D.2.在正項(xiàng)等比數(shù)列{an}中,a2,a48是方程2x2-7x+6=0的兩個(gè)根,那么a1·a2·a25·a48·a49的值為()A. B.9 C.±9 D.353.設(shè)首項(xiàng)為1,公比為的等比數(shù)列{an}的前n項(xiàng)和為Sn,那么()A.Sn=2an-1 B.Sn=3an-2C.Sn=4-3an D.Sn=3-2an4.{an}為等比數(shù)列,a4+a7=2,a5a6=-8,那么a1+a10=A.7 B.5 C.-5 D.-75.等差數(shù)列{an}的公差為2,假設(shè)a2,a4,a8成等比數(shù)列,那么{an}的前n項(xiàng)和Sn=()A.n(n+1) B.n(n-1)C. D.6.設(shè)數(shù)列{an}是首項(xiàng)為a1,公差為-1的等差數(shù)列,Sn為其前n項(xiàng)和.假設(shè)S1,S2,S4成等比數(shù)列,那么a1的值為.

7.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,假設(shè)S2=4,an+1=2Sn+1,n∈N*,那么a1=,S5=.

8.(2023江蘇,9)等比數(shù)列{an}的各項(xiàng)均為實(shí)數(shù),其前n項(xiàng)和為Sn.S3=,S6=,那么a8=.

9.{an}是公差為3的等差數(shù)列,數(shù)列{bn}滿足b1=1,b2=,anbn+1+bn+1=nbn.(1)求{an}的通項(xiàng)公式;(2)求{bn}的前n項(xiàng)和.10.等差數(shù)列{an}的前n項(xiàng)和為Sn,且S4=4(a3+1),3a3=5a4,數(shù)列{bn}是等比數(shù)列,且b1b2=b3,2b1=a(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)求數(shù)列{|an|}的前n項(xiàng)和Tn.11.在數(shù)列{an}中,Sn為數(shù)列{an}的前n項(xiàng)和,且Sn=1+kan(k≠0,且k≠1).(1)求通項(xiàng)公式an;(2)當(dāng)k=-1時(shí),求+…+的值.能力提升12.(2023四川廣元二診)數(shù)列{an}的前n項(xiàng)和為Sn,且對(duì)任意正整數(shù)n都有an=Sn+2成立.假設(shè)bn=log2an,那么b1008=()A.2017 B.2016 C.2015 D.201413.假設(shè)a,b是函數(shù)f(x)=x2-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),且a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,那么p+q的值等于()A.6 B.7 C.8 D.914.設(shè)等比數(shù)列{an}滿足a1+a3=10,a2+a4=5,那么a1a2…an的最大值為15.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn.S2=4,an+1=2Sn+1,n∈N*.(1)求通項(xiàng)公式an;(2)求數(shù)列{|an-n-2|}的前n項(xiàng)和.高考預(yù)測(cè)16.數(shù)列{an}滿足a1=5,a2=5,an+1=an+6an-1(n≥2).(1)求證:{an+1+2an}是等比數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.答案：1.C解析:∵a3a5=4(a4-1),∴=4(a4-1),解得a4=2又a4=a1q3,且a1=,∴q=2.∴a2=a1q=.2.B解析:∵a2,a48是方程2x2-7x+6=0的兩個(gè)根,∴a2·a48=3.又a1·a49=a2·a48==3,a25>0,∴a1·a2·a25·a48·a49==9.選B.3.D解析:Sn==3-2an,應(yīng)選D.4.D解析:∵{an}為等比數(shù)列,∴a5a6=a4a7=-聯(lián)立可解得當(dāng)時(shí),q3=-,故a1+a10=+a7q3=-7;當(dāng)時(shí),q3=-2,故a1+a10=+a7q3=-7.綜上可知,a1+a10=-7.5.A解析:∵a2,a4,a8成等比數(shù)列,∴=a2·a8,即(a1+6)2=(a1+2)(a1+14),解得a1=2.∴Sn=na1+d=2n+n2-n=n2+n=n(n+1).應(yīng)選A.6.-解析:由得S1=a1,S2=a1+a2=2a1-1,S4=4a1+×(-1)=4a而S1,S2,S4成等比數(shù)列,∴(2a1-1)2=a1(4a1整理,得2a1+1=0,解得a17.1121解析:由題意,可得a1+a2=4,a2=2a1+所以a1=1,a2=3.再由an+1=2Sn+1,an=2Sn-1+1(n≥2),得an+1-an=2an,即an+1=3an(n≥2).又因?yàn)閍2=3a1,所以數(shù)列{an}是以1為首項(xiàng),3為公比的等比數(shù)列所以S5==121.8.32解析:設(shè)該等比數(shù)列的公比為q,那么S6-S3==14,即a4+a5+a6=14.①∵S3=,∴a1+a2+a3=.由①得(a1+a2+a3)q3=14,∴q3==8,即q=2.∴a1+2a1+4a1=,a1∴a8=a1·q7=×27=32.9.解:(1)由,a1b2+b2=b1,b1=1,b2=,得a1=2.所以數(shù)列{an}是首項(xiàng)為2,公差為3的等差數(shù)列,通項(xiàng)公式為an=3n-1.(2)由(1)和anbn+1+bn+1=nbn得bn+1=,因此{(lán)bn}是首項(xiàng)為1,公比為的等比數(shù)列.記{bn}的前n項(xiàng)和為Sn,那么Sn=.10.解:(1)設(shè)等差數(shù)列{an}的公差為d.∵S4=4(a3+1),3a3=5a∴解得∴an=11-2n.設(shè)數(shù)列{bn}的公比為q.∵b1b2=b3,2b1=a5,∴解得∴bn=.(2)由(1)知,Sn=10n-n2.由an=11-2n≤0可知n≥5.5,即a1>0,a2>0,…,a5>0,a6<0,a7<0,…,an<0.故當(dāng)n≤5時(shí),Tn=Sn=10n-n2;當(dāng)n≥6時(shí),Tn=2S5-Sn=n2-10n+50.于是Tn=11.解:(1)∵S1=a1=1+ka1,∴a1=.又an=Sn-Sn-1(n≥2),∴an=an-1(n≥2).∴an==-.(2)∵在數(shù)列{an}中,a1=,q=,∴{}是首項(xiàng)為,公比為的等比數(shù)列.當(dāng)k=-1時(shí),等比數(shù)列{}的首項(xiàng)為,公比為,∴+…+.12.A解析:在an=Sn+2中,令n=1得a1=8.∵an=Sn+2成立,∴an+1=Sn+1+2成立,兩式相減得an+1-an=an+1,∴an+1=4an,又a1≠0,∴數(shù)列{an}為等比數(shù)列,∴an=8·4n-1=22n+1,∴bn=log2an=2n+1,∴b1008=2017,應(yīng)選A.13.D解析:∵a,b是函數(shù)f(x)=x2-px+q(p>0,q>0)的兩個(gè)不同的零點(diǎn),∴a+b=p,ab=q.∵p>0,q>0,∴a>0,b>0.又a,b,-2這三個(gè)數(shù)可適當(dāng)排序后成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,∴①或②.解①得解②得∴p=a+b=5,q=1×4=4.∴p+q=9.應(yīng)選D.14.64解析:由a1+a3=10,a2+a4=a1q+a3q=5,兩式相除得,解得q=,a1=8,所以a1a2…an=8n·,拋物線f(n)=-n2+n的對(duì)稱軸為n=-=3.又n∈N*,所以當(dāng)n=3或n=4時(shí),a1a2…an取最大值為=26=6415.解:(1)由題意得又當(dāng)n≥2時(shí),由an+1-an=(2Sn+1)-(2Sn-1+1)=2an,得an+1=3an.所以,數(shù)列{an}的通項(xiàng)公式為an=3n-1,n∈N*.(2)設(shè)bn=|3n-1-n-2|,n∈N*,b1=2,b2=1.當(dāng)n≥3時(shí),由于3n-1>n+2,故bn=3n-1-n-2,n≥3.設(shè)數(shù)列{bn}的前n項(xiàng)和為Tn,那么T1=2,T2=3.當(dāng)n≥3時(shí),Tn=3+,所以Tn=16.(1)證明:∵an+1=an+6an-1(n≥2),∴an+1+2an=3an+6an-1=3(an+2an-1)(n≥2).又a1=5,a2=5,∴a2+2a1=∴an+2an-1≠0(n≥2),∴=3(n≥2),∴數(shù)列{an+1+2an}是以15為首項(xiàng),3