4.3等比數(shù)列4.3.1等比數(shù)列的概念第1課時(shí)等比數(shù)列的概念及通項(xiàng)公式必備知識(shí)基礎(chǔ)練1.(2021北京豐臺(tái)高二期末)已知等比數(shù)列{an}滿(mǎn)足a1=1,a4=8,則a7等于()A.32 B.32 C.64 D.642.(2021天津河西高二期末)數(shù)列1,22,12,24A.12n1 B.22nC.(1)n22n1 D.(1)n+122n13.(2021江蘇啟東高二期末)在等比數(shù)列{an}中,a5a2=4,a4a1=2,則公比q=()A.±12 B.±2 C.12 D4.在等比數(shù)列{an}中,a1a2=2,a2a4=16,則公比q等于()A.2 B.3 C.48 D.25.(多選題)設(shè){an}為等比數(shù)列,給出四個(gè)數(shù)列:①{2an};②{an2};③{2an};④{log2|an|}.A.① B.② C.③ D.④6.在160與5中間插入4個(gè)數(shù),使它們同這兩個(gè)數(shù)成等比數(shù)列,則這4個(gè)數(shù)依次為.

7.在數(shù)列{an}中,已知a1=3,且對(duì)任意正整數(shù)n都有2an+1an=0,則an=.

8.在等比數(shù)列{an}中,若a1=18,公比q=2,則a4與a8的等比中項(xiàng)是.9.已知數(shù)列{an}是等差數(shù)列,且a2=3,a4+3a5=56.若log2bn=an.(1)求證:數(shù)列{bn}是等比數(shù)列;(2)求數(shù)列{bn}的通項(xiàng)公式.10.已知數(shù)列{an}滿(mǎn)足a1=32,且an+1=λan+1n∈N*,λ∈R且λ≠23.求使數(shù)列{an+1}是等比數(shù)列的λ的值.11.(2021湖北黃岡中學(xué)高三模擬)已知在數(shù)列{an}中,a1=1且2an+1=6an+2n1(n∈N*).(1)求證:數(shù)列an+n2為等比數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.關(guān)鍵能力提升練12.已知數(shù)列{an}是等比數(shù)列,則方程組a1x+aA.唯一解 B.無(wú)解C.無(wú)數(shù)多組解 D.不能確定13.(2021江蘇常州高二期中)數(shù)列{an}中,a1=12,am+n=aman(?m,n∈N*),則a6=(A.116 B.132 C.16414.(2021湖南長(zhǎng)沙四校高二聯(lián)考)在數(shù)列{an}中,對(duì)任意n∈N*,都有an+12an=0,則2a1+aA.14 B.13 C.1215.(2021廣東廣州高二期末)在等比數(shù)列{an}中,a1+a2+a3=6,a4+a5+a6=3,則a7+a8+a9=()A.24 B.32 C.34 D16.(多選題)已知數(shù)列{an},{bn}是等比數(shù)列,那么下列一定是等比數(shù)列的是()A.{k·an} B.1C.{an+bn} D.{an·bn}17.(多選題)(2021江蘇蘇州高二期中)已知{an}為等比數(shù)列,下列結(jié)論正確的是()A.若a3=2,則a22+B.a32+C.若a3=a5,則a1=a2 D.若a5>a3,則a7>a518.已知一個(gè)等比數(shù)列的各項(xiàng)均為正數(shù),且它的任何一項(xiàng)都等于它后兩項(xiàng)的和,則它的公比q=.

19.若數(shù)列a1,a2a1,a3a2,…,anan-1,20.(2021安徽亳州高二期末)已知數(shù)列{an}滿(mǎn)足a1=12,an+1=an2-an,若bn=1an1,則數(shù)列{bn21.已知數(shù)列{an}的前n項(xiàng)和Sn=2an+1,(1)求證:{an}是等比數(shù)列,并求出其通項(xiàng)公式;(2)設(shè)bn=an+1+2an,求證:數(shù)列{bn}是等比數(shù)列.22.已知數(shù)列{an}和{bn}滿(mǎn)足:a1=λ,an+1=23an+n4,bn=(1)n·(an3n+21),其中λ為實(shí)數(shù),n為正整數(shù)(1)對(duì)任意實(shí)數(shù)λ,證明數(shù)列{an}不是等比數(shù)列;(2)試判斷數(shù)列{bn}是否為等比數(shù)列,并證明你的結(jié)論.學(xué)科素養(yǎng)創(chuàng)新練23.(多選題)(2022湖北鄂州高二期中)在數(shù)列{an}中,如果對(duì)任意n∈N*都有an+2-an+1an+1-an=k(k為常數(shù)),則稱(chēng){an}A.等差數(shù)列一定是等差比數(shù)列B.等差比數(shù)列的公差比一定不為0C.若an=3n+2,則數(shù)列{an}是等差比數(shù)列D.若等比數(shù)列是等差比數(shù)列,則其公比等于公差比24.已知數(shù)列{cn},其中cn=2n+3n,數(shù)列{cn+1pcn}為等比數(shù)列,求常數(shù)p.參考答案4.3等比數(shù)列4.3.1等比數(shù)列的概念第1課時(shí)等比數(shù)列的概念及通項(xiàng)公式1.D設(shè)等比數(shù)列{an}的公比為q,則a4=a1q3=q3=8,解得q=2,故a7=a1q6=64.2.D根據(jù)數(shù)列的項(xiàng)可知該數(shù)列是一個(gè)以1為首項(xiàng),22為公比的等比數(shù)列,所以該數(shù)列的通項(xiàng)公式為1×22n1=(1)n1×22n1=(1)n+1×22n1.3.D由a1q4.A在等比數(shù)列{an}中,a1a2=2,a2a4=16,則a2a4a1a2=q3=5.AB設(shè)等比數(shù)列{an}的公比為q,則2an2an-1=aan2an-12=a取等比數(shù)列an=(1)n,則{2an}的前三項(xiàng)為12,2,12,不成等比數(shù)列;此時(shí)log2|an|=0,{log2|an故選AB.6.80,40,20,10設(shè)這6個(gè)數(shù)所成等比數(shù)列的公比為q,則5=160q5,∴q5=132,∴q=1∴這4個(gè)數(shù)依次為80,40,20,10.7.3·12n-1由2an+1an=0,得an+1an=12因?yàn)閍1=3,所以an=3·128.±4依題意,得a6=a1q5=18×25=4,而a4與a8的等比中項(xiàng)是±a6,故a4與a8的等比中項(xiàng)是±49.(1)證明由log2bn=an,得bn=2a因?yàn)閿?shù)列{an}是等差數(shù)列,不妨設(shè)公差為d,則bnbn-1=2an2an-1=2an-(2)解由已知,得a解得a1=-1,d=4,于是b1=21=12,公比q=2d=24=16,所以數(shù)列{bn}的通項(xiàng)公式10.解若數(shù)列{an+1}是等比數(shù)列,則an+1+1an+1即(λμ)an+2μ=0,對(duì)于任意n∈N*恒成立,則λ-μ=0,2故當(dāng)λ=2時(shí),數(shù)列{an+1}是等比數(shù)列.11.(1)證明∵2an+1=6an+2n1(n∈N*),∴an+1=3an+n12∴an+1+∵a1+12=1+1∴an+n2為等比數(shù)列,首項(xiàng)為32,公比為3.(2)解由(1)得,an+n2=32×3n1=1∴an=12×3nn12.C由題意,數(shù)列{an}是等比數(shù)列,可得a1a4=a2a5=a3a6,所以直線(xiàn)a1所以方程組a1x13.C由于?m,n∈N*,有am+n=aman,且a1=12令m=1,則an+1=a1an=12an,即數(shù)列{an}是首項(xiàng)為12,公比為1所以an=12×12n1=12n,故a6=126=164.14.A由an+12an=0得an+1an=2,即數(shù)列{an}則2a15.B設(shè)等比數(shù)列{an}的公比為q,則a4+a5+a6=q3(a1+a2+a3),即6q3=3,可得q3=12,因此a7+a8+a9=q3(a4+a5+a6)=12×(3)=3216.BD由題意,可設(shè)等比數(shù)列{an}的公比為q1(q1≠0),則an=a1·q1n-1,等比數(shù)列{bn}的公比為q2(q2≠0),則bn=b1·q2n-1,對(duì)于A,當(dāng)k=0時(shí),{k·a對(duì)于B,1a∴數(shù)列1an是一個(gè)以1a1為首項(xiàng),1q1對(duì)于C,舉出反例,當(dāng)an=1,bn=1時(shí),數(shù)列{an+bn}不是等比數(shù)列,故C錯(cuò)誤;對(duì)于D,an·bn=a1·b1(q1·q2)n1,∴數(shù)列{an·bn}是一個(gè)以a1b1為首項(xiàng),q1q2為公比的等比數(shù)列,故D正確.故選BD.17.ABD若a3=2,則a22+a42≥2a2a4=2a32=8,當(dāng)a2=a4=±2因?yàn)閍32+a52≥2a3a5=2a42,當(dāng)a3=a5設(shè)等比數(shù)列的公比為q,因?yàn)閍3=a5,所以q2=a5a3=1,所以q=±1,當(dāng)q=1時(shí),a1=a2,故設(shè)等比數(shù)列的公比為q,則q2>0,因?yàn)閍5>a3,所以a5q2>a3q2,即a7>a5,故D正確.故選ABD.18.-1+52依題意,得an=an+1+a所以an=anq+anq2.因?yàn)閍n>0,所以q2+q1=0,解得q=-1+19.32由題意,得anan-1=(2)n1(n≥2),所以a2a1=2,a3a2=(2)2,a4a3=(2)3,a5a4=(2)4又a1=1,所以a5=32.20.2n1因?yàn)閍n+1=an2-an,所以1an+1=2an1,所以1an+11=2an2=2所以數(shù)列{bn}是首項(xiàng)為1,公比為2的等比數(shù)列,所以bn=1×2n1=2n1.21.證明(1)∵Sn=2an+1,∴Sn+1=2an+1+1,Sn+1Sn=an+1=(2an+1+1)(2an+1)=2an+12an,∴an+1=2an.由已知及上式可知an≠0.∴由an+1an=2知{a由a1=S1=2a1+1,得a1=1,∴an=2n1.(2)由(1)知,an=2n1,∴bn=an+1+2an=2n2×2n1=2×2n=2n+1=4×2n1.bn+1bn=-4×2n-22.(1)證明假設(shè)存在一個(gè)實(shí)數(shù)λ,使{an}是等比數(shù)列,則有a22=a1a3,即23λ32=λ49λ4?49λ24λ+9=49λ24λ?9=0,所以{an}不是等比數(shù)列.(2)解是等比數(shù)列,證明如下:因?yàn)閎n+1=(1)n+1·[an+13(n+1)+21]=(1)n+1·23an2n+14=23(1)n·(an3n+21)=23bn又b1=(λ+18),所以當(dāng)λ=18時(shí),bn=0(n∈N*),此時(shí){bn}不是等比數(shù)列;當(dāng)λ≠18時(shí),b1=(λ+18)≠0,由上可知bn≠0,所以bn+1bn=23(n故當(dāng)λ≠18時(shí),數(shù)列{bn}是以(λ+18)為首項(xiàng),23為公比的等比數(shù)列23.BCD對(duì)于等差數(shù)列{an},考慮an=1,an+1=1,an+2=1,an+2-an+1an+1-an無(wú)意義,所以A選項(xiàng)錯(cuò)誤;若等差比數(shù)列的公差比為0,an+2-an+1an+1-an=0,an+2an+1=