1、培优课等比数列习题课必备知识基础练1.等比数列an中,a1+a3=10,a4+a6=54,则数列an的通项公式为()A.an=24-nB.an=2n-4C.an=2n-3D.an=23-n2.已知数列an的前n项和Sn=2n-1+1,则数列an的前10项中所有奇数项之和与所有偶数项之和的比为()A.12B.2C.172341D.3411723.(2022陕西宝鸡二模)已知数列an是公比为q的等比数列,若2a1=a3a4,且a5是a4与2的等差中项,则q的值是()A.1B.2C.-1或1D.-2或24.在等比数列an中,a1+a2=6,a3+a4=12,则数列an的前8项和为()A.90B.30
2、(2+1)C.45(2+1)D.725.等差数列an的前n项和为Sn,S5=15,S9=18,在等比数列bn中,b3=a3,b5=a5,则b7的值为()A.23B.43C.2D.36.(多选题)已知等差数列an的首项为1,公差d=4,前n项和为Sn,则下列结论成立的有()A.数列Snn的前10项和为100B.若a1,a3,am成等比数列,则m=21C.若i=1n1aiai+1625,则n的最小值为6D.若am+an=a2+a10,则1m+16n的最小值为25127.数列an的前n项和为Sn,且a1=1,2Sn=Sn+1,则a2+a4+a2n=.8.已知各项不为0的等差数列an满足2a2-a62
3、+2a10=0,首项为18的等比数列bn的前n项和为Sn,若b6=a6,则S6=.9.已知数列an,用a1,a2,a3,an,构造一个新数列a1,(a2-a1),(a3-a2),(an-an-1),此数列是首项为1,公比为13的等比数列.求:(1)数列an的通项;(2)数列an的前n项和Sn.关键能力提升练10.已知等比数列an的前n项和为Sn,且a1+a3=52,a2+a4=54,则Snan=()A.4n-1B.4n-1C.2n-1D.2n-111.数列an中,a1=2,am+n=aman.若ak+1+ak+2+ak+10=215-25,则k=()A.2B.3C.4D.512.如图,已知点D
4、为ABC的边BC上一点,BD=3DC,En(nN+)为边AC上的动点,满足EnA=14an+1EnB-(3an+2)EnD.在数列an中,an0,a1=1,则数列an的通项公式为()A.an=32n-1-2B.an=2n-1C.an=3n-2D.an=23n-1-113.数列an的前n项和为Sn,已知a1=15,且对任意正整数m,n,都有am+n=aman,若Sn1,令bn=an+1(n=1,2,),若数列bn有连续四项在集合-53,-23,19,37,82中,则6q=.15.等比数列an的前n项和为Sn,已知S1,S3,S2成等差数列.(1)求数列an的公比q;(2)若a1-a3=3,求Sn
5、.16.设an是公比不为1的等比数列,a1为a2,a3的等差中项.(1)求an的公比;(2)若a1=1,求数列nan的前n项和.17.设数列an满足:a1=1,an+1=3an.设Sn为数列bn的前n项和,已知b10,2bn-b1=S1Sn.(1)求数列an,bn的通项公式;(2)设cn=bnlog3an,求数列cn的前n项和Tn;(3)证明对任意nN+且n2,有1a2-b2+1a3-b3+1an-bn625,解得n6,故n的最小值为7,故选项C错误;由等差数列的性质可知m+n=12,所以1m+16n=1121m+16n(m+n)=1121+nm+16mn+16112(17+24)=2512,
6、当且仅当nm=16mn时,即n=4m=485时取等号,因为m,nN+,所以等号不成立,故选项D错误.故选AB.7.4n-13由2Sn=Sn+1,可得2Sn=Sn+an+1,所以an+1=Sn,则当n2时,an+1-an=Sn-Sn-1=an,即an+1=2an,所以数列an从第2项开始是公比为2的等比数列,且a2=S1=a1=1,所以an=2n-2(n2),则数列a2,a4,a2n是首项为1,公比为4的等比数列,则a2+a4+a2n=1-4n1-4=4n-13.8.6382a2-a62+2a10=0,4a6=a62.a60,a6=4,b6=4.又数列bn的首项b1=18,设其公比为q,q5=b
7、6b1=32,q=2,S6=18-421-2=638.9.解(1)an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=321-13n.(2)Sn=a1+a2+a3+an=321-13+321-132+321-133+321-13n=32n-131+13+132+13n-1=32n-121-13n1-13=32n-341-13n=34(2n-1)+1413n-1.10.D设数列an的公比为q,a1+a3=52,a2+a4=54,a1+a1q2=52,a1q+a1q3=54,得1+q2q+q3=2,解得q=12.代入得a1=2,an=2
8、12n-1=42n,Sn=21-12n1-12=41-12n,Snan=41-12n42n=2n-1,故选D.11.Cam+n=aman,令m=1,又a1=2,an+1=a1an=2an,an+1an=2,an是以2为首项,2为公比的等比数列,an=2n.ak+1+ak+2+ak+10=2k+1+2k+2+2k+10=2k+11-2101-2=2k+11-2k+1=215-25.k+11=15,k+1=5,解得k=4.12.D因为BD=3DC,EnC=EnB+4DC,又DC=DEn+EnC,所以EnC=-13EnB+43EnD.设mEnC=EnA,则由EnA=14an+1EnB-(3an+2)
9、EnD,得-13m=14an+1,43m=-(3an+2),所以14an+1=14(3an+2),所以an+1+1=3(an+1).因为a1+1=2,所以数列an+1是以2为首项,3为公比的等比数列,所以an+1=23n-1,所以an=23n-1-1.故选D.13.14令m=1,则an+1an=a1,由a1=15,知an+1=15an,数列an是以a1为首项,15为公比的等比数列.an=15n,Sn=15-15n+11-15=141-15n=14145n14.由SnSn的最大值,可知t的最小值为14.14.-9由题意知,数列bn有连续四项在集合-53,-23,19,37,82中,说明an有连续
10、四项在集合-54,-24,18,36,81中,由于an中连续四项至少有一项为负数,q1,数列an的连续四项为-24,36,-54,81,q=36-24=-32,6q=-9.15.解(1)由题意得a1+(a1+a1q)=2(a1+a1q+a1q2),a10,2q2+q=0.又q0,q=-12.(2)由已知可得a1-a1-122=3,故a1=4,Sn=41-12n1-12=831-12n.16.解(1)设an的公比为q,由题设得2a1=a2+a3,即2a1=a1q+a1q2.所以q2+q-2=0,解得q=1(舍去),q=-2.故an的公比为-2.(2)记Sn为nan的前n项和.由(1)及题设可得,
11、an=(-2)n-1.所以Sn=1+2(-2)+n(-2)n-1,-2Sn=-2+2(-2)2+(n-1)(-2)n-1+n(-2)n.可得3Sn=1+(-2)+(-2)2+(-2)n-1-n(-2)n=1-(-2)n3-n(-2)n.所以Sn=19(3n+1)(-2)n9.17.(1)解an+1=3an,数列an是公比为3,首项为1的等比数列,an=3n-1.2bn-b1=S1Sn,当n=1时,2b1-b1=S1S1,S1=b1,b10,b1=1.当n2时,bn=Sn-Sn-1=2bn-2bn-1,bn=2bn-1,数列bn是公比为2,首项为1的等比数列,bn=2n-1.(2)解cn=bnl
12、og3an=2n-1log33n-1=(n-1)2n-1,Tn=020+121+222+(n-2)2n-2+(n-1)2n-1,2Tn=021+122+223+(n-2)2n-1+(n-1)2n,-得-Tn=020+21+22+23+2n-1-(n-1)2n=2n-2-(n-1)2n=-2-(n-2)2n,Tn=(n-2)2n+2.(3)证明当n2时,1an-bn=13n-1-2n-1=133n-2-2n-1=13n-2+2(3n-2-2n-2)13n-2,1a2-b2+1a3-b3+1an-bn130+131+13n-2=1-13n-11-13=321-13n-132.18.(1)解当n=2
13、时,4S4+5S2=8S3+S1,即41+32+54+a4+51+32=81+32+54+1,解得a4=78.(2)证明4Sn+2+5Sn=8Sn+1+Sn-1(n2),4Sn+2-4Sn+1+Sn-Sn-1=4Sn+1-4Sn(n2),即4an+2+an=4an+1(n2).当n=1时,4a3+a1=454+1=6=4a2,当n=1时,4an+2+an=4an+1成立.an+2-12an+1an+1-12an=4an+2-2an+14an+1-2an=4an+1-an-2an+14an+1-2an=2an+1-an2(2an+1-an)=12,数列an+1-12an是以a2-12a1=1为首项,以12为公比的等比数列.(3)解由(2),知an+1-12an=12n-1,即an+112n+1an12n=4,数列an12n是以a112=2为首项,以4为公差的等差数列,an12n=2+(n-1)4=4n-2,即an=(4n-2)12n=(2n-1)12n-1,数列an的通项公式是an=(2n-1)12n-1.
