（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 第六章 數(shù)列 考點(diǎn)規(guī)范練29 等比數(shù)列及其前n項(xiàng)和.docx

考點(diǎn)規(guī)范練29等比數(shù)列及其前n項(xiàng)和基礎(chǔ)鞏固組1.(2018北京高考)“十二平均律”是通用的音律體系,明代朱載堉最早用數(shù)學(xué)方法計(jì)算出半音比例,為這個(gè)理論的發(fā)展做出了重要貢獻(xiàn).十二平均律將一個(gè)純八度音程分成十二份,依次得到十三個(gè)單音,從第二個(gè)單音起,每一個(gè)單音的頻率與它的前一個(gè)單音的頻率的比都等于122.若第一個(gè)單音的頻率為f,則第八個(gè)單音的頻率為()A.32fB.322fC.1225fD.1227f答案D解析設(shè)第n個(gè)單音的頻率為an,由題意,anan-1=122(n2),所以an為等比數(shù)列,因?yàn)閍1=f,所以a8=a1(122)7=1227f,故選D.2.已知an是等比數(shù)列,則“a2<a4”是“an是單調(diào)遞增數(shù)列”的()A.充分不必要條件B.必要不充分條件C.充分必要條件D.既不充分也不必要條件答案B解析在等比數(shù)列-1,2,-4,8,中,滿足a2<a4,但an是單調(diào)遞增數(shù)列不成立,即充分性不成立,若an是單調(diào)遞增數(shù)列,則必有a2<a4,即必要性成立,則“a2<a4”是“an是單調(diào)遞增數(shù)列”的必要不充分條件.故選B.3.已知一個(gè)蜂巢里有1只蜜蜂,第1天,它飛出去找回了4個(gè)伙伴;第2天,5只蜜蜂飛出去,各自找回了4個(gè)伙伴,按照這個(gè)規(guī)律繼續(xù)下去,第20天所有的蜜蜂都?xì)w巢后,蜂巢中一共有蜜蜂()A.420只B.520只C.520-54只D.421-43只答案B解析由題意,可設(shè)蜂巢里的蜜蜂數(shù)為數(shù)列an,則a1=1+4=5,a2=54+5=25,an=5an-1,故數(shù)列an為等比數(shù)列,首項(xiàng)a1=5,公比q=5,故第20天所有的蜜蜂都?xì)w巢后,蜂巢中一共有a20=5519=520只蜜蜂.4.設(shè)實(shí)數(shù)列an,bn分別為等差數(shù)列與等比數(shù)列,且a1=b1=4,a4=b4=1,則以下結(jié)論正確的是()A.a2>b2B.a3<b3C.a5>b5D.a6>b6答案A解析a1=4,a4=1,d=-1.b1=4,b4=1,又0<q<1,q=2-23,b2=243<a2=3,b3<223<a3=2,b5=2-23>a5=0,b6=2-43>a6=-1.5.數(shù)列an滿足an+1=an-1(nN*,R,且0),若數(shù)列an-1是等比數(shù)列,則的值等于()A.1B.-1C.12D.2答案D解析由an+1=an-1,得an+1-1=an-2=an-2.由an-1是等比數(shù)列,所以2=1,得=2.6.等比數(shù)列an的前n項(xiàng)和為Sn,已知a1=1,a1,S2,5成等差數(shù)列,則數(shù)列an的公比q=.答案2解析由題意得2S2=a1+5,即2(1+q)=1+5,q=2.7.在各項(xiàng)均為正數(shù)的等比數(shù)列an中,a3=2-1,a5=2+1,則a32+2a2a6+a3a7=.答案8解析由等比數(shù)列性質(zhì),得a3a7=a52,a2a6=a3a5,所以a32+2a2a6+a3a7=a32+2a3a5+a52=(a3+a5)2=(2-1+2+1)2=(22)2=8.8.已知各項(xiàng)都為正數(shù)的數(shù)列an滿足a1=1,an2-(2an+1-1)an-2an+1=0,則a3=,an=.答案1412n-1解析由題意得a2=12,a3=14.(等比數(shù)列的定義、通項(xiàng)公式)由an2-(2an+1-1)an-2an+1=0得2an+1(an+1)=an(an+1).因?yàn)閍n的各項(xiàng)都為正數(shù),所以an+1an=12.故an是首項(xiàng)為1,公比為12的等比數(shù)列,因此an=12n-1.能力提升組9.已知數(shù)列an是等比數(shù)列,a2=2,a5=14,則a1a2+a2a3+anan+1=()A.16(1-4-n)B.16(1-2-n)C.323(1-4-n)D.323(1-2-n)答案C解析由a5=14=a2q3=2q3,解得q=12,可知數(shù)列anan+1仍是等比數(shù)列:其首項(xiàng)是a1a2=8,公比為14.所以a1a2+a2a3+anan+1=81-14n1-14=323(1-4-n).10.已知等比數(shù)列an的前n項(xiàng)和為Sn,a1+a3=30,S4=120,設(shè)bn=1+log3an,則數(shù)列bn的前15項(xiàng)和為()A.152B.135C.80D.16答案B解析由題設(shè)可得a2+a4=S4-(a1+a3)=90,即q(a1+a3)=90q=3,所以a1=301+9=3,則an=33n-1=3n.所以bn=1+log3(3n)=1+n,則數(shù)列bn是首項(xiàng)為b1=2,公差為d=1的等差數(shù)列.所以S15=215+15142=135,應(yīng)選B.11.已知數(shù)列an滿足a1=1,an+1an=2n,則S2 015=()A.22 015-1B.21 009-3C.321 007-3D.21 008-3答案B解析a1=1,an+1an=2n,an0,a2=2,當(dāng)n2時(shí),anan-1=2n-1.an+1an-1=2n2n-1=2(n2).數(shù)列an中奇數(shù)項(xiàng),偶數(shù)項(xiàng)分別成等比數(shù)列.S2015=1-210081-2+2(1-21007)1-2=21009-3.故選B.12.(2018浙江高考)已知a1,a2,a3,a4成等比數(shù)列,且a1+a2+a3+a4=ln(a1+a2+a3).若a1>1,則()A.a1<a3,a2<a4B.a1>a3,a2<a4C.a1<a3,a2>a4D.a1>a3,a2>a4答案B解析設(shè)等比數(shù)列的公比為q,則a1+a2+a3+a4=a1(1-q4)1-q,a1+a2+a3=a1(1-q3)1-q.a1+a2+a3+a4=ln(a1+a2+a3),a1+a2+a3=ea1+a2+a3+a4,即a1(1+q+q2)=ea1(1+q+q2+q3).又a1>1,q<0.假設(shè)1+q+q2>1,即q+q2>0,解得q<-1(q>0舍去).由a1>1,可知a1(1+q+q2)>1,a1(1+q+q2+q3)>0,即1+q+q2+q3>0,即(1+q)+q2(1+q)>0,即(1+q)(1+q2)>0,這與q<-1相矛盾.1+q+q2<1,即-1<q<0.a1>a3,a2<a4.13.已知a,b為實(shí)常數(shù),ci(iN*)是公比不為1的等比數(shù)列,直線ax+by+ci=0與拋物線y2=2px(p>0)均相交,所成弦的中點(diǎn)為Mi(xi,yi),則下列說(shuō)法錯(cuò)誤的是()A.數(shù)列xi可能是等比數(shù)列B.數(shù)列yi是常數(shù)列C.數(shù)列xi可能是等差數(shù)列D.數(shù)列xi+yi可能是等比數(shù)列答案C解析由直線ax+by+ci=0,當(dāng)a=0,b0時(shí),直線by+ci=0與拋物線y2=2px(p>0)僅有一個(gè)交點(diǎn),不合題意.當(dāng)a0,b=0時(shí),直線ax+ci=0,化為x=-cia,則xi=-cia,yi=0,xi+yi=-cia.由ci(iN*)是公比不為1的等比數(shù)列,可得xi是等比數(shù)列,xi+yi是等比數(shù)列,不是等差數(shù)列.當(dāng)a0,b0時(shí),直線ax+by+ci=0化為x=-bay-cia,代入拋物線y2=2px(p>0),可得y2+2pbay+2pcia=0.根據(jù)根與系數(shù)的關(guān)系可得Mi:pb2a2-cia,-pba.yi是常數(shù)列,是等比數(shù)列,也是等差數(shù)列.綜上可得:A,B,D都有可能,只有C不可能.故選C.14.如圖,在等腰直角三角形ABC中,斜邊BC=22,過(guò)點(diǎn)A作BC的垂線,垂足為A1;過(guò)點(diǎn)A1作AC的垂線,垂足為A2;過(guò)點(diǎn)A2作A1C的垂線,垂足為A3;,依此類推,設(shè)BA=a1,AA1=a2,A1A2=a3,A5A6=a7,則a7=.答案14解析由題意知數(shù)列an是以首項(xiàng)a1=2,公比q=22的等比數(shù)列,a7=a1q6=2226=14.15.已知數(shù)列an的前m(m4)項(xiàng)是公差為2的等差數(shù)列,從第m-1項(xiàng)起,am-1,am,am+1,成公比為2的等比數(shù)列.若a1=-2,則m=,an的前6項(xiàng)和S6=.答案428解析因?yàn)閍m-1=a1+(m-2)d=2m-6,am=2m-4,而2m-42m-6=2,解得m=4,所以數(shù)列an的前6項(xiàng)依次為-2,0,2,4,8,16.所以S6=28.16.(2018浙江溫嶺模擬)已知數(shù)列an,a1=1,an+1=2an-n2+3n(nN*),若新數(shù)列an+n2+n是等比數(shù)列,則=,=.答案-11解析an+1=2an-n2+3n可化為an+1+(n+1)2+(n+1)=2(an+n2+n),即an+1=2an+n2+(-2)n-,=-1,-2=3,-=0,解得=-1,=1.an+1=2an-n2+3n可化為an+1-(n+1)2+(n+1)=2(an-n2+n).又a1-12+10,故=-1,=1時(shí)可使得數(shù)列an+n2+n是等比數(shù)列.17.已知正項(xiàng)數(shù)列an的奇數(shù)項(xiàng)a1,a3,a5,a2k-1,構(gòu)成首項(xiàng)a1=1的等差數(shù)列,偶數(shù)項(xiàng)構(gòu)成公比q=2的等比數(shù)列,且a1,a2,a3成等比數(shù)列,a4,a5,a7成等差數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=a2n+1a2n,Tn=b1b2bn,求正整數(shù)k,使得對(duì)任意nN*,均有TkTn.解(1)由題意得a22=a1a3,2a5=a4+a7,設(shè)a1,a3,a5,a2k-1,的公差為d,則a3=1+d,a5=1+2d,a7=1+3d,a4=2a2,代入a22=1(1+d),1+d=2a2,又a2>0,解得a2=2,d=3.故數(shù)列an的通項(xiàng)公式為an=3n-12,n為奇數(shù),2n2,n為偶數(shù),(2)bn=3n+12n,顯然bn>0,bn+1bn=3n+42n+13n+12n=3n+46n+2<1,數(shù)列bn單調(diào)遞減.又b1=2,b2=74,b3=108,b4=136,b1>b2>b3>1>b4>b5>.當(dāng)k=3時(shí),對(duì)任意nN*,均有T3Tn.18.(2014浙江高考)已知數(shù)列an和bn滿足a1a2a3an=(2)bn(nN*).若an為等比數(shù)列,且a1=2,b3=6+b2.(1)求an與bn;(2)設(shè)cn=1an-1bn(nN*).記數(shù)列cn的前n項(xiàng)和為Sn.求Sn;求正整數(shù)k,使得對(duì)任意nN*均有SkSn.解(1)由題意a1a2a3an=(2)bn,b3-b2=6,知a3=(2)b3-b2=8,又由a1=2,得公比q=2(q=-2,舍去),所以數(shù)列an的通項(xiàng)為an=2n(nN*).所以,a1a2a3an=2n(n+1)2=(2)n(n+1).故數(shù)列bn的通項(xiàng)為bn=n(n+1)(nN*).(2)由(1)知cn=1an-1bn=12n-1n-1n+1(nN*),所以Sn=1n+1-12n(nN*).因?yàn)閏1=0,c2>0,c3>0,c4>0,當(dāng)n5時(shí),cn=1n(n+1)n(n+1)2n-1,而n(n+1)2n-(n+1)(n+2)2n+1=(n+1)(n-2)2n+1>0,得n(n+1)2n5(5+1)25<1.所以,當(dāng)n5時(shí),cn<0.綜上,對(duì)任意nN*恒有S4Sn,故k=4.