新高考數(shù)學(xué)一輪復(fù)習(xí)專題六數(shù)列6-3等比數(shù)列練習(xí)含答案

6.3等比數(shù)列五年高考考點(diǎn)1等比數(shù)列及其前n項(xiàng)和1.(2023全國(guó)甲理,5,5分,中)設(shè)等比數(shù)列{an}的各項(xiàng)均為正數(shù),前n項(xiàng)和為Sn,若a1=1,S5=5S3-4,則S4=(C)A.158B.658C.152.(2022全國(guó)乙,文10,理8,5分,中)已知等比數(shù)列{an}的前3項(xiàng)和為168,a2-a5=42,則a6=(D)A.14B.12C.6D.33.(2020課標(biāo)Ⅱ理,6,5分,中)數(shù)列{an}中,a1=2,am+n=aman.若ak+1+ak+2+…+ak+10=215-25,則k=(C)A.2B.3C.4D.5(2023全國(guó)甲文,13,5分,易)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若8S6=7S3,則{an}的公比為-125.(2023全國(guó)乙理,15,5分,中)已知{an}為等比數(shù)列,a2a4a5=a3a6,a9a10=-8,則a7=-2.

6.(2023北京,14,5分,中)我國(guó)度量衡的發(fā)展有著悠久的歷史,戰(zhàn)國(guó)時(shí)期就已經(jīng)出現(xiàn)了類似于砝碼的、用來(lái)測(cè)量物體質(zhì)量的“環(huán)權(quán)”.已知9枚環(huán)權(quán)的質(zhì)量(單位:銖)從小到大構(gòu)成項(xiàng)數(shù)為9的數(shù)列{an},該數(shù)列的前3項(xiàng)成等差數(shù)列,后7項(xiàng)成等比數(shù)列,且a1=1,a5=12,a9=192,則a7=48;數(shù)列{an}所有項(xiàng)的和為384.

7.(2020新高考Ⅱ,18,12分,易)已知公比大于1的等比數(shù)列{an}滿足a2+a4=20,a3=8.(1)求{an}的通項(xiàng)公式;(2)求a1a2-a2a3+…+(-1)n-1anan+1.解析(1)設(shè)公比為q(q>1),依題意有a解得a1=2,q=2或a所以an=2n,所以數(shù)列{an}的通項(xiàng)公式為an=2n.(2)由(1)知(-1)n-1anan+1=(-1)n-1×2n×2n+1=(-1)n-122n+1,所以a1a2-a2a3+…+(-1)n-1anan+1=23-25+27-29+…+(-1)n-122n+1=23[1?(?22)n]1?(?228.(2022新高考Ⅱ,17,10分,中)已知{an}是等差數(shù)列,{bn}是公比為2的等比數(shù)列,且a2-b2=a3-b3=b4-a4.(1)證明:a1=b1;(2)求集合{k|bk=am+a1,1≤m≤500}中元素的個(gè)數(shù).解析(1)證明:設(shè)等差數(shù)列{an}的公差為d.由a2-b2=a3-b3得a1+d-2b1=a1+2d-4b1,故d=2b1,①由a3-b3=b4-a4得a1+2d-4b1=8b1-a1-3d,故2a1+5d=12b1,②由①②得2a1+10b1=12b1,即a1=b1.(2)由(1)知d=2b1=2a1,由bk=am+a1,1≤m≤500得b1×2k-1=2a1+(m-1)d,即a1×2k-1=2a1+2(m-1)a1,其中a1≠0,∴2k-1=2m,即2k-2=m,∴1≤2k-2≤500,∴0≤k-2≤8,∴2≤k≤10.故集合{k|bk=am+a1,1≤m≤500}中元素的個(gè)數(shù)為9.考點(diǎn)2等比數(shù)列的性質(zhì)及其應(yīng)用1.(2021全國(guó)甲文,9,5分,中)記Sn為等比數(shù)列{an}的前n項(xiàng)和.若S2=4,S4=6,則S6=(A)A.7B.8C.9D.102.(2023新課標(biāo)Ⅱ,8,5分,中)記Sn為等比數(shù)列{an}的前n項(xiàng)和,若S4=-5,S6=21S2,則S8=(C)A.120B.85C.-85D.-120

三年模擬練速度1.(2024山東青島第一次適應(yīng)性檢測(cè),1)等比數(shù)列{an}中,a2=1,a5=8,則a7=(A)A.32B.24C.20D.162.(2024廣東惠州一模,2)設(shè)正項(xiàng)等比數(shù)列{an}的公比為q,若a2,3a1,a3成等差數(shù)列,則q=(B)A.12B.2C.133.(2024安徽蚌埠教學(xué)質(zhì)量檢查,4)已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,若a5=9,則log3a4+log3a6=(C)A.2B.3C.4D.94.(2024山東新高考聯(lián)合質(zhì)量測(cè)評(píng),3)已知數(shù)列{an}是等比數(shù)列,{bn}是等差數(shù)列,a1=1,若b1,2a2,3a3,2b3為常數(shù)列,則a4b2=(C)A.0B.8C.827D.5.(2024湖南九校聯(lián)盟第二次聯(lián)考,2)已知{an}是等比數(shù)列,Sn是其前n項(xiàng)和.若a3-a1=3,S4=5S2,則a2的值為(C)A.2B.4C.±2D.±46.(2024廣東江門一模,3)已知{an}是等比數(shù)列,a3a5=8a4,且a2,a6是方程x2-34x+m=0兩根,則m=(C)A.8B.-8C.64D.-647.(2024山東濰坊一模,4)已知數(shù)列{an}滿足a1=0,a2=1.若數(shù)列{an+an+1}是公比為2的等比數(shù)列,則a2024=(A)A.22023+13C.21012-1D.21011-18.(多選)(2024湖南長(zhǎng)沙雅禮中學(xué)二模,10)在《增刪算法統(tǒng)宗》中有這樣一則故事:“三百七十八里關(guān),初行健步不為難;次日腳痛減一半,如此六日過(guò)其關(guān).”則下列說(shuō)法正確的是(ACD)A.此人第二天走了九十六里路B.此人第三天走的路程占全程的1C.此人第一天走的路程比后五天走的路程多六里D.此人后三天共走了四十二里路9.(2024山東淄博一模,13)已知等比數(shù)列{an}共有2n+1項(xiàng),a1=1,所有奇數(shù)項(xiàng)的和為85,所有偶數(shù)項(xiàng)的和為42,則公比q=2.

10.(2024湖北八市聯(lián)考,13)設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,若3S2>S6>0,則公比q的取值范圍為(-1,0)∪(0,1).

11.(2024江蘇南京、鹽城調(diào)研,17)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,an+Sn=1.(1)求數(shù)列{an}的通項(xiàng)公式;(2)數(shù)列{bn}滿足anbn=cosnπ2,求{bn}的前50項(xiàng)和T解析(1)由an+Sn=1,得an-1+Sn-1=1(n≥2),兩式相減得an-an-1+an=0(n≥2),即an=12an-1(n≥2當(dāng)n=1時(shí),2S1=2a1=1,得a1=12≠0,所以anan?1=故{an}是首項(xiàng)為12,公比為12的等比數(shù)列,從而an=(2)由(1)得bn=2ncosnπ所以T50=-22+24-26+28-…-250=?4[1?(?4)25]1?(?4)=-4512.(2024河北唐山一模,15)已知數(shù)列{an}是正項(xiàng)等比數(shù)列,其前n項(xiàng)和為Sn,且a2a4=16,S5=S3+24.(1)求{an}的通項(xiàng)公式;(2)記{an+log2an}的前n項(xiàng)和為Tn,求滿足Tn<2024的最大整數(shù)n.解析(1)設(shè){an}的公比為q,則an=a1qn-1,因?yàn)閍n>0,所以q>0,依題意可得a3=4,整理得q2+q-6=0,解得q=2或q=-3(舍去),所以an=a3qn-3=2n-1.(2)由(1)可知an+log2an=2n-1+n-1,故Tn=(20+21+22+…+2n-1)+(0+1+2+…+n-1)=2n-1+n(顯然,Tn隨著n的增大而增大,T10=210-1+45=1068<2024,T11=211-1+55=2102>2024,所以滿足Tn<2024的最大整數(shù)n=10.13.(2024浙江寧波二模,16)已知等差數(shù)列{an}的公差為2,記數(shù)列{bn}的前n項(xiàng)和為Sn,b1=0,b2=2且滿足bn+1=2Sn+an.(1)證明:數(shù)列{bn+1}是等比數(shù)列;(2)求數(shù)列{anbn}的前n項(xiàng)和Tn.解析(1)證明:n≥2時(shí),bn+1-bn=2(Sn-Sn-1)+an-an-1=2bn+2,即bn+1=3bn+2.又b1=0,b2=2,所以n≥1時(shí),bn+1=3bn+2,即bn+1+1=3(bn+1).又b1+1=1≠0,所以bn+1≠0,所以bn+1+1bn+1=3.所以數(shù)列{(2)由(1)易得bn=3n-1-1.由b2=2b1+a1可得a1=2,所以an=2n.所以anbn=2n(3n-1-1)=2n·3n-1-2n,所Tn=2(1·30+2·31+3·32+…+n·3n-1)-n(n+1).令M=1·30+2·31+3·32+…+n·3n-1,則3M=1·31+2·32+3·33+…+n·3n,所以2M=-(30+31+32+…+3n-1)+n·3n=n·3n-1?3n1?3所以Tn=2M-n(n+1)=(2n?1)3n+12-練思維1.(多選)(2024湖南長(zhǎng)沙適應(yīng)性考試,12)設(shè)等比數(shù)列{an}的公比為q,前n項(xiàng)積為Tn,下列說(shuō)法正確的是(BCD)A.若T8=T12,則a10a11=1B.若T8=T12,則T20=1C.若a1=1024,且T10為數(shù)列{Tn}的唯一最大項(xiàng),則12109<D.若a1>0,且T10>T11>T9,則使得Tn>1成立的n的最大值為202.(2024江西贛州二模,17)已知數(shù)列{an}滿足a1=14,an,32an+1,2anan+1(1)求證:數(shù)列1an?1是等比數(shù)列,并求出{an(2)記{an}的前n項(xiàng)和為Sn,證明:381-13n≤Sn<5解析(1)由于an,32an+1,2anan+1成等差數(shù)列所以3an+1=an+2anan+1,(1分)即1an+1=3an-2,可得1an+1所以1an?1是以1a1-1=3為首項(xiàng),3為公比的等比數(shù)列,所以1an-1=3×3n-1=3n,即an=13n+1.(2)證明:因?yàn)?3n+1所以Sn=a1+a2+a3+…+an<a1+132+1=14+132+133+…+13n=14+191?13n?1由于381?13n=14·1?13n1?13=141+所以Sn=a1+a2+a3+…+an≥14130+13=14×1?13n1?13=381?13故381?13n≤Sn<5123.(2024黑龍江哈師大附中三模,16)已知數(shù)列{an}的前n項(xiàng)和為Sn,Sn=3an-2n(n∈N*).(1)求證:數(shù)列{an-2n}是等比數(shù)列;(2)設(shè)bn=an+λ·2n-(λ+1)·32n?1,若{bn}是遞增數(shù)列,求實(shí)數(shù)解析(1)證明:由Sn=3an-2n,得Sn+1=3an+1-2n+1,兩式相減得2an+1=3an+2n.(2分)2(an+1-2n+1)=3(an-2n),即an+1-2n+1=32(an-2n),(4分)∵S1=3a1-21,∴a1=1,∴a1-21=-1≠0,∴an-2n≠0.(6分)∴an+1?∴{an-2n}是以-1為首項(xiàng),32為公比的等比數(shù)列.(7分)(2)由(1)可知an-2n=(-1)·32n?1,∴an=2n-32n?1∴bn=(1+λ)2n-(λ+2)32則bn+1=(1+λ)2n+1-(λ+2)32∵{bn}是遞增數(shù)列,∴bn+1>bn對(duì)任意的n∈N*恒成立,∴(1+λ)2n+1-(λ+2)32n>(1+λ)2n-(λ+2)∴3(1+λ)>(λ+2)34n,(10分當(dāng)λ+2<0,即λ<-2時(shí),3(1+λ)λ∵34n>0,且n→+∞時(shí),34n→0,∴-2<λ≤-1(舍).(12分)當(dāng)λ+2=0,即λ=-2時(shí),-3>0,矛盾,故λ=-2(舍).(13分)當(dāng)λ+2>0,即λ>-2時(shí),3(1+λ)λ∵34n≤341=34,∴λ>-23,滿足λ>-2,故λ>-23.(15分4.(2024江蘇南通二模,18)已知數(shù)列{an}的前n項(xiàng)和為Sn,Sn=an-4an+1,a1=-1.(1)證明:數(shù)列{2an+1-an}為等比數(shù)列;(2)設(shè)bn=an+4n(n+1),求數(shù)列{b(3)是否存在正整數(shù)p,q(p<6<q),使得Sp,S6,Sq成等差數(shù)列?若存在,求p,q;若不存在,說(shuō)明理由.解析(1)證明:當(dāng)n=1時(shí),S1=a1-4a2,故a2=0.當(dāng)n≥2時(shí),由Sn=an-4an+1,①得Sn-1=an-1-4an,②①-②得,an=an-an-1-4an+1+4an,即4an+1=4an-an-1.(2分)所以2(2an+1-an)=2an-an-1.又2a2-a1=1≠0,所以?n∈N*,2an+1-an≠0,所以2an+1所以{2an+1-an}是以1為首項(xiàng),12為公比的等比數(shù)列.(4分)(2)由(1)得,2an+1-an=12所以2nan+1-2n-1an=1,則2a2-a1=1,4a3-2a2=1,……,2n-1an-2n-2an-1=1,通過(guò)累加法可得2n-1an=n-2,解得an=n?22n?1,(所以bn=n+22n所以{bn}的前n項(xiàng)和為181?12×2+12×2?122×3+…+12n?1(3)由(1)知,Sn=n?22n?1-4×因?yàn)镾p,S6,Sq成等差數(shù)列,所以-1225=-p2整理得p2p+q2q=316.因?yàn)閜2p+q2q=316,又1≤p<6,p∈N*,經(jīng)檢驗(yàn),只能p=5.(14分)所以532+q2q=316,故令dn=n2n,則dn+1-dn=1?n2n+1≤0,所以d1=d2>d3>d因?yàn)閐8=132,所以q=8所以存在p=5,q=8,使得Sp,S6,Sq成等差數(shù)列.(17分)練風(fēng)向1.(新定義理解)(多選)(2024山東煙臺(tái)、德州診斷,11)給定數(shù)列{an},定義差分運(yùn)算:Δan=an+1-an,Δ2an=Δan+1-Δan,n∈N*.若數(shù)列{an}滿足an=n2+n,數(shù)列{bn}的首項(xiàng)為1,且Δbn=(n+2)·2n-1,n∈N*,則(BC)A.存在M>0,使得Δan<M恒成立B.存在M>0,使得Δ2an<M恒成立C.對(duì)任意M>0,總存在n∈N*,使得bn>MD.對(duì)任意M>0,總存在n∈N*,使得Δ2b2.(創(chuàng)新知識(shí)交匯)(2024廣東二模,18)已知正項(xiàng)數(shù)列{an},{bn},滿足an+1=bn+c2,bn+1=an+(1)若a1≠b1,且a1+b1≠2c,證明:數(shù)列{an-bn}和{an+bn-2c}均為等比數(shù)列;(2)若a1>b1,a1+b1=2c,以an,bn,c為三角形三邊長(zhǎng)構(gòu)造序列△AnBnCn(其中AnBn=c,BnCn=an,AnCn=bn),記△AnBnCn外接圓的面積為Sn,證明:Sn>π3c2(3)在(2)的條件下證明:數(shù)列{Sn}是遞減數(shù)列.證明(1)由an+1=bn+c2,bn+1=an+c2,兩式相減得an+1-bn+1=-12(an-因?yàn)閍1≠b1,所以a1-b1≠0,(2分)所以{an-bn}