福建专用2022高考数学一轮复习课时规范练30等比数列及其前n项和理新人教A版

课时规范练30　等比数列及其前n项和一、基础巩固组1.已知等比数列{an}满足a1=14,a3a5=4(a4-1),则a2=(　　)A.2B.1C.12D.182.在正项等比数列{an}中,a2,a48是方程2x2-7x+6=0的两个根,则a1·a2·a25·a48·a49的值为(　　)A.212B.93C.±93D.353.(2022安徽黄山市二模,理3)已知数列{an}的前n项和为Sn,且a1=2,an+1=Sn+1(n∈N*),则S5=(　　)A.31B.42C.37D.474.设首项为1,公比为23的等比数列{an}的前n项和为Sn,则(　　)A.Sn=2an-1B.Sn=3an-2C.Sn=4-3anD.Sn=3-2an5.(2022全国Ⅲ,理9)等差数列{an}的首项为1,公差不为0.若a2,a3,a6成等比数列,则{an}前6项的和为(　　)A.-24B.-3C.3D.86.(2022辽宁鞍山一模,理4)已知数列{an}满足an2=an-1·an+1(n≥2),若a2=3,a2+a4+a6=21,则a4+a6+a8=(　　)A.84B.63C.42D.21〚导学号21500732〛7.设数列{an}是首项为a1,公差为-1的等差数列,Sn为其前n项和.若S1,S2,S4成等比数列,则a1的值为　　　　　. 8.(2022北京,理10)若等差数列{an}和等比数列{bn}满足a1=b1=-1,a4=b4=8,则a2b2=　　　　　. 9.(2022江苏,9)等比数列{an}的各项均为实数,其前n项和为Sn.已知S3=74,S6=634,则a8=　　　　　. 10.(2022安徽池州模拟)设数列{an}的前n项和为Sn,a1=1,且数列{Sn}是以2为公比的等比数列.(1)求数列{an}的通项公式;(2)求a1+a3+…+a2n+1.二、综合提升组11.(2022四川广元二诊,理6)已知数列{an}的前n项和为Sn,且对任意正整数n都有an=34Sn+2成立.若bn=log2an,则b1008=(　　)A.2017B.2016C.2015D.201412.(2022河南南阳期末,理5)已知各项均为正数的等比数列{an},a3·a5=2,若f(x)=x(x-a1)(x-a2)·…·(x-a7),则f'(0)=(　　)A.82B.-82C.128D.-12813.已知{an}是公差为3的等差数列,数列{bn}满足b1=1,b2=13,anbn+1+bn+1=nbn.(1)求{an}的通项公式;(2)求{bn}的前n项和.三、创新应用组14.已知数列{an}的前n项和Sn满足Sn=2an+(-1)n.(1)求数列{an}的前三项a1,a2,a3;(2)求证:数列an+23(-1)n为等比数列,并求出{an}的通项公式.3\n〚导学号21500733〛课时规范练30　等比数列及其前n项和1.C　∵a3a5=4(a4-1),∴a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,∴q=2,∴a2=a1q=12.2.B　∵a2,a48是方程2x2-7x+6=0的两个根,∴a2·a48=3.又a1·a49=a2·a48=a252=3,a25>0,∴a1·a2·a25·a48·a49=a255=93.3.D　∵an+1=Sn+1(n∈N*),∴Sn+1-Sn=Sn+1(n∈N*),∴Sn+1+1=2(Sn+1)(n∈N*),∴数列{Sn+1}是首项为3,公比为2的等比数列.则S5+1=3×24,解得S5=47.4.D　Sn=a1(1-qn)1-q=a1-anq1-q=1-23an1-23=3-2an,故选D.5.A　设等差数列的公差为d,则d≠0,a32=a2·a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=6×1+6×52×(-2)=-24,故选A.6.C　∵an2=an-1·an+1(n≥2),∴数列{an}是等比数列,设其公比为q,∵a2=3,a2+a4+a6=3+3q2+3q4=21,即q4+q2-6=0,解得q2=2或q2=-3(舍去),∴a4+a6+a8=a2q2+a4q2+a6q2=2(a2+a4+a6)=42,故选C.7.-12　由已知得S1=a1,S2=a1+a2=2a1-1,S4=4a1+4×32×(-1)=4a1-6,而S1,S2,S4成等比数列,∴(2a1-1)2=a1(4a1-6),整理,得2a1+1=0,解得a1=-12.8.1　设等差数列{an}的公差为d,等比数列{bn}的公比为q,由题意知-1+3d=-q3=8,即-1+3d=8,-q3=8,解得d=3,q=-2.故a2b2=-1+3-1×(-2)=1.9.32　设该等比数列的公比为q,则S6-S3=634-74=14,即a4+a5+a6=14.①∵S3=74,∴a1+a2+a3=74.由①得(a1+a2+a3)q3=14,∴q3=1474=8,即q=2.∴a1+2a1+4a1=74,a1=14,∴a8=a1·q7=14×27=32.10.解(1)∵S1=a1=1,且数列{Sn}是以2为公比的等比数列,∴Sn=2n-1,又当n≥2时,an=Sn-Sn-1=2n-2(2-1)=2n-2.当n=1时,a1=1,不适合上式.∴an=1,n=1,2n-2,n≥2.(2)a3,a5,…,a2n+1是以2为首项,4为公比的等比数列,∴a3+a5+…+a2n+1=2(1-4n)1-4=2(4n-1)3.∴a1+a3+…+a2n+1=1+2(4n-1)3=22n+1+13.11.A　在an=34Sn+2中,令n=1得a1=8,∵an=34Sn+2成立,3\n∴an+1=34Sn+1+2成立,两式相减得an+1-an=34an+1,∴an+1=4an,又a1≠0,∴数列{an}为等比数列,∴an=8·4n-1=22n+1,∴bn=log2an=2n+1,∴b1008=2017,故选A.12.B13.解(1)由已知,得a1b2+b2=b1,因为b1=1,b2=13,所以a1=2.所以数列{an}是首项为2,公差为3的等差数列,通项公式为an=3n-1.(2)由(1)和anbn+1+bn+1=nbn,得bn+1=bn3,因此{bn}是首项为1,公比为13的等比数列.记{bn}的前n项和为Sn,则Sn=1-13n1-13=32-12×3n-1.14.(1)解在Sn=2an+(-1)n中分别令n=1,2,3,得a1=2a1-1,a1+a2=2a2+1,a1+a2+a3=2a3-1,解得a1=1,a2=0,a3=2.(2)证明由Sn=2an+(-1)n(n∈N*)得Sn-1=2an-1+(-1)n-1(n≥2),两式相减,得an=2an-1-2(-1)n(n≥2).∴an=2an-1-43(-1)n-23(-1)n=2an-1+43(-1)n-1-23(-1)n(n≥2),∴an+23(-1)n=2an-1+23(-1)n-1(n≥2).∴数列an+23(-1)n是以a1-23=13为首项,以2为公比的等比数列.∴an+23(-1)n=13×2n-1.∴an=2n-13-23(-1)n.3