2024-2025學(xué)年新教材高中數(shù)學(xué)第四章數(shù)列4.3.1第1課時(shí)等比數(shù)列的概念及通項(xiàng)公式課后提升訓(xùn)練含解析新人教A版選擇性必修第二冊(cè)

第四章數(shù)列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)公式課后篇鞏固提升基礎(chǔ)達(dá)標(biāo)練1.等比數(shù)列{an}中,a1a2=2,a2a4=16,則公比q等于()A.2 B.3 C.48 D.2解析等比數(shù)列{an}中,a1a2=2,a2a4=16,∴a2·a4a1·a2=q3=8,答案A2.(多選)(2024福建廈門一中高一月考)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|},其中肯定A.① B.② C.③ D.④解析設(shè)等比數(shù)列{an}的公比為q,則2an2an-1=aan2a故{an2}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.答案AB3.已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前4項(xiàng)的和為15,且a5=3a3+4a1,則a3=()A.16 B.8 C.4 D.2解析設(shè)該等比數(shù)列的首項(xiàng)為a1,公比為q,由已知得,a1q4=3a1q2+4a1,因?yàn)閍1>0且q>0,則可得q=2,又因?yàn)閍1(1+q+q2+q3)=15,即可解得a1=1,則a3=a1q2=4.答案C4.已知等差數(shù)列{an}的公差為2,若a1,a3,a4成等比數(shù)列,則a2=()A.-4 B.-6 C.-8 D.-10解析∵a4=a1+6,a3=a1+4,a1,a3,a4成等比數(shù)列,∴a32=a1·a4,即(a1+4)2=a1·(a1解得a1=-8,∴a2=a1+2=-6.故選B.答案B5.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=1,Sn=2an+1,則Sn=()A.2n-1 B.32n-1 C.解析由Sn=2an+1,得Sn=2(Sn+1-Sn),即2Sn+1=3Sn,Sn+1Sn=32.又S1=a1=1,所以S答案B6.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.

解析設(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.答案80,40,20,107.在數(shù)列{an}中,已知a1=3,且對(duì)隨意正整數(shù)n都有2an+1-an=0,則an=.

解析由2an+1-an=0,得an+1an=12,所以數(shù)列{an}是等比數(shù)列,公比為12.因?yàn)閍1=3,答案3·18.在等比數(shù)列{an}中,若a1=18,q=2,則a4與a8的等比中項(xiàng)是.解析依題意,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±4答案±49.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.(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=2-1=12,公比q=2d=24=16,所以數(shù)列{bn}的通項(xiàng)公式10.已知數(shù)列{an}滿意a1=1,an+1=2an+1.(1)證明:數(shù)列{an+1}是等比數(shù)列;(2)求數(shù)列{an}的通項(xiàng)公式.(1)證明∵an+1=2an+1,∴an+1+1=2(an+1).由a1=1,知a1+1≠0,從而an+1≠0.∴an+1+1an+1=2(∴數(shù)列{an+1}是等比數(shù)列.(2)解由(1)知{an+1}是以a1+1=2為首項(xiàng),2為公比的等比數(shù)列.∴an+1=2·2n-1=2n.即an=2n-1.實(shí)力提升練1.若a,b,c成等差數(shù)列,而a+1,b,c和a,b,c+2都分別成等比數(shù)列,則b的值為()A.16 B.15 C.14 D.12解析依題意,得2b=答案D2.在等比數(shù)列{an}中,a1=1,公比|q|≠1.若am=a1a2a3a4a5,則m等于()A.9 B.10 C.11 D.12解析∵am=a1a2a3a4a5=q·q2·q3·q4=q10=1×q10,∴m=11.答案C3.(多選)(2024山東滕州第一中學(xué)新校高二月考)已知數(shù)列{an},{bn}是等比數(shù)列,那么下列肯定是等比數(shù)列的是()A.{k·an} B.1C.{an+bn} D.{an·bn}解析由題意,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=b對(duì)于A,當(dāng)k=0時(shí),{k·an}明顯不是等比數(shù)列,故A錯(cuò)誤;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)n-1,∴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.答案BD4.已知-7,a1,a2,-1四個(gè)實(shí)數(shù)成等差數(shù)列,-4,b1,b2,b3,-1五個(gè)實(shí)數(shù)成等比數(shù)列,則a2-a1解析由題意,得a2-a1=-1-(-7)3=2,b22=(-4)×(-1)=4.又b2是等比數(shù)列中的第3項(xiàng),所以b2與第1項(xiàng)同號(hào),即b2答案-15.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)都等于它的后面兩項(xiàng)的和,則它的公比q=.

解析依題意,得an=an+1+an+2,所以an=anq+anq2.因?yàn)閍n>0,所以q2+q-1=0,解得q=-1+答案-6.若數(shù)列a1,a2a1,a3a2,…,anan-1,…是首項(xiàng)為解析由題意,得anan-1=(-2)n-1(n≥2),所以a2a1=-2,a3a2=(-2)2,a4a3=(-2)3,a5a4=(-2)4,將上面的四個(gè)式子兩邊分別相乘,得a5a1=(-2答案327.已知各項(xiàng)都為正數(shù)的數(shù)列{an}滿意a1=1,an2-(2an+1-1)an-2an+1=(1)求a2,a3;(2)求{an}的通項(xiàng)公式.解(1)由題意可得a2=12,a3=1(2)由an2-(2an+1-1)an-2an+1=0,得2an+1(an+1)=an(an+1).因?yàn)閧an}的各項(xiàng)都為正數(shù),所以an+1an=12.故{因此an=12n-1,n8.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.證明(1)∵Sn=2an+1,∴Sn+1=2an+1+1,Sn+1-Sn=an+1=(2an+1+1)-(2an+1)=2an+1-2an,∴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=-2n-1.(2)由(1)知,an=-2n-1,∴bn=an+1+2an=-2n-2×2n-1=-2×2n=-2n+1=-4×2n-1.bn+1bn=-4×2n-素養(yǎng)培優(yōu)練已知數(shù)列{cn},其中cn=2n+3n,數(shù)列{cn+1-pcn}為等比數(shù)列,求常數(shù)p.解因?yàn)閿?shù)列{cn+1-pcn}為等比