2020版高考數(shù)學(xué)一輪復(fù)習(xí) 第六章 數(shù)列 課時(shí)規(guī)范練29 等比數(shù)列及其前n項(xiàng)和 文 北師大版.doc

課時(shí)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北京師大附中期中)在等比數(shù)列an中,a1=3,a1+a2+a3=9,則a4+a5+a6等于()A.9B.72C.9或72D.9或-722.(2018湖南岳陽(yáng)一中期末)等比數(shù)列an中,anan+1=4n-1,則數(shù)列an的公比為()A.2或-2B.4C.2D.23.(2018黑龍江仿真模擬十一)等比數(shù)列an中,an>0,a1+a2=6,a3=8,則a6=()A.64B.128C.256D.5124.在公比為正數(shù)的等比數(shù)列an中,a1+a2=2,a3+a4=8,則S8等于()A.21B.42C.135D.1705.(2018重慶梁平二調(diào))我國(guó)古代數(shù)學(xué)名著算法統(tǒng)宗中有如下問(wèn)題:“遠(yuǎn)望巍巍塔七層,紅光點(diǎn)點(diǎn)倍加增,共燈三百八十一,請(qǐng)問(wèn)尖頭幾盞燈?”意思是:一座7層塔共掛了381盞燈,且相鄰兩層中的下一層燈數(shù)是上一層燈數(shù)的2倍,則塔的頂層共有燈()A.1盞B.3盞C.5盞D.9盞6.(2018衡水中學(xué)仿真,6)已知數(shù)列an為等比數(shù)列,且a2a3a4=-a72=-64,則tana4a63=()A.-3B.3C.3D.-337.(2018陜西咸陽(yáng)三模)已知數(shù)列an為等比數(shù)列,且a3a11+2a72=4,則tan(a1a13)的值為.8.(2018全國(guó)3,文17)等比數(shù)列an中,a1=1,a5=4a3.(1)求an的通項(xiàng)公式;(2)記Sn為an的前n項(xiàng)和,若Sm=63,求m.9.(2018北京城六區(qū)一模)已知等比數(shù)列an滿足以a1=1,a5=a2.(1)求數(shù)列an的通項(xiàng)公式;(2)試判斷是否存在正整數(shù)n,使得an的前n項(xiàng)和Sn為?若存在,求出n的值;若不存在,說(shuō)明理由.綜合提升組10.(2018河南六市聯(lián)考一,10)若正項(xiàng)遞增等比數(shù)列an滿足1+(a2-a4)+(a3-a5)=0(R),則a6+a7的最小值為()A.-2B.-4C.2D.411.(2018全國(guó)1,理14)記Sn為數(shù)列an的前n項(xiàng)和.若Sn=2an+1,則S6=.12.已知數(shù)列an的前n項(xiàng)和為Sn,對(duì)任意的正整數(shù)n,都有Sn=an+n-3成立.求證:存在實(shí)數(shù),使得數(shù)列an+為等比數(shù)列.13.已知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.創(chuàng)新應(yīng)用組14.(2018浙江,10)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>a415.我們把滿足xn+1=xn-f(xn)f(xn)的數(shù)列xn叫做牛頓數(shù)列.已知函數(shù)f(x)=x2-1,數(shù)列xn為牛頓數(shù)列,設(shè)an=lnxn-1xn+1,已知a1=2,則a3=.課時(shí)規(guī)范練29等比數(shù)列及其前n項(xiàng)和1.D設(shè)等比數(shù)列an的公比為q,a1=3,a1+a2+a3=9,3+3q+3q2=9,解得q=1或q=-2,當(dāng)q=1時(shí),a4+a5+a6=(a1+a2+a3)q3=9.當(dāng)q=-2時(shí),a4+a5+a6=-72,故選D.2.C設(shè)等比數(shù)列an的公比為q,anan+1=4n-1>0,an+1an+2=4n且q>0,兩式相除可得an+1an+2anan+1=4n4n-1=4,即q2=4,q=2,故選C.3.A由題意結(jié)合等比數(shù)列的通項(xiàng)公式可得a1+a1q=6,a1q2=8,a1>0,解得a1=2,q=2,則a6=a1q5=225=64.4.D(方法一)S8=(a1+a2)+(a3+a4)+(a5+a6)+(a7+a8)=2+8+32+128=170.(方法二)q2=a3+a4a1+a2=4,又q>0,q=2,a1(1+q)=a1(1+2)=2,a1=23,S8=23(28-1)2-1=170.5.B設(shè)塔的頂層共有x盞燈,則各層的燈數(shù)構(gòu)成一個(gè)公比為2的等比數(shù)列,由x(1-27)1-2=381,可得x=3,故選B.6.A依題意,得a2a3a4=a33=-64,所以a3=-4.由a72=64,得a7=-8,或a7=8(由于a7與a3同號(hào),故舍去),所以a4a6=a3a7=32.tana4a63=tan323=tan11-=-tan=-3,故選A.7.3an是等比數(shù)列,a3a11+2a72=a72+2a72=4,即a72=43,a1a13=a72=43,tan(a1a13)=tan43=3.8.解 (1)設(shè)an的公比為q,由題設(shè)得an=qn-1.由已知得q4=4q2,解得q=0(舍去),q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,則Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程沒(méi)有正整數(shù)解.若an=2n-1,則Sn=2n-1.由Sm=63得2m=64,解得m=6.綜上,m=6.9.解 (1)設(shè)an的公比為q,a5=a2,且a5=a2q3,q3=18,得q=12,an=a1qn-1=12n-1(n=1,2,).(2)不存在n,使得an的前n項(xiàng)和Sn為52,a1=1,q=12,Sn=1-12n1-12=21-12n.(方法一)令Sn=52,則21-12n=52,得2n=-4,該方程無(wú)解,不存在n,使得an的前n項(xiàng)和Sn為52.(方法二)對(duì)任意nN+,有1-12n<1,Sn=21-12n<2,不存在n,使an的前n項(xiàng)和Sn為52.10.D因?yàn)?+(a2-a4)+(a3-a5)=0,所以1+q=1a4-a2(q>1),a6+a7=a6(1+q)=a6a4-a2=q4q2-1=(q2-1+1)2q2-1=(q2-1)+2+1q2-12+2(q2-1)1q2-1=4,當(dāng)且僅當(dāng)q=2時(shí)取等號(hào),即a6+a7的最小值為4,故選D.11.-63Sn=2an+1,Sn-1=2an-1+1(n2).-,得an=2an-2an-1,即an=2an-1(n2).又S1=2a1+1,a1=-1.an是以-1為首項(xiàng),2為公比的等比數(shù)列,則S6=-1(1-26)1-2=-63.12.證明 Sn=an+n-3,當(dāng)n=1時(shí),S1=32a1+1-3,所以a1=4.當(dāng)n2時(shí),Sn-1=32an-1+n-1-3,由兩式相減得an=32an-32an-1+1,即an=3an-1-2(n2).變形得an-1=3(an-1-1),而a1-1=3,數(shù)列an-1是首項(xiàng)為3,公比為3的等比數(shù)列,存在實(shí)數(shù)=-1,使得數(shù)列an-1為等比數(shù)列.13.解 (1)由已知,得a1b2+b2=b1,因?yàn)閎1=1,b2=,所以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=bn3,因此bn是首項(xiàng)為1,公比為13的等比數(shù)列.記bn的前n項(xiàng)和為Sn,則Sn=1-13n1-13=32-123n-1.14.B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<a4.15.8由f(x)=x2-1,得f(x)=2x,則xn+1=xn-xn2-12xn=xn2+12xn,所以xn+1-1=(xn-1)22xn,xn+1+1=(xn+1)22xn,所以xn+1-1xn+1+1=(xn-1)2(xn+1)2,所以lnxn+1-1xn+1+1=ln(xn-1)2(xn+1)2=2lnxn-1xn+1,即an+1=2an,所以數(shù)列an是首項(xiàng)為2,公比為2的等比數(shù)列,則a3=222=8.