文科数学一轮复习高考帮全国版试题第6章第1讲数列（考题帮数学文）

第一讲数列的概念与简单表示法题组数列的通项公式及前n项和1.[2016浙江,13,6分]设数列{an}的前n项和为Sn.若S2=4,an+1=2Sn+1,n∈N*,则a1=,S5=.

2.[2015江苏,11,5分][文]设数列{an}满足a1=1,且an+1an=n+1(n∈N*),则数列{1an}前10项的和为3.[2014新课标全国Ⅱ,16,5分][文]数列{an}满足an+1=11-an,a8=2,则a14.[2013新课标全国Ⅰ,14,5分]若数列{an}的前n项和Sn=23an+13,则{an}的通项公式是an=5.[2016全国卷Ⅲ,17,12分][文]已知各项都为正数的数列{an}满a1=1,an2(2an+11)an2an+1=(Ⅰ)求a2,a3;(Ⅱ)求{an}的通项公式.6.[2015新课标全国Ⅰ,17,12分]Sn为数列{an}的前n项和,已知an>0,an2+2an=4Sn+(Ⅰ)求{an}的通项公式;(Ⅱ)设bn=1anan+1,求数列{bA组基础题1.[2017云南省11校调考,3]在数列{an}中,a1=3,an+1=3anan+3,则a4A.34 B.1 C.43 2.[2017贵州省高招适应性考试,3]已知数列{an}满足an=12an+1,若a3+a4=2,则a4+a5=()A.12 B.1 C.4 3.[2017昆明市高三质检,5]已知数列{an}的前n项和为Sn,且2,Sn,an成等差数列,则S17=()A.0 B.2 C.2 D.344.[2018惠州市二调,15]已知数列{an}满足a1=1,an+12an=2n(n∈N*),则数列{an}的通项公式an=.

5.[2017郑州市第三次质量预测,14]若数列{an}的前n项和为Sn,且3Sn2an=1,则{an}的通项公式是an=.

6.[2018南昌市摸底调研,17]已知数列{an}的前n项和Sn=2n+12,记bn=anSn(n∈N*).(1)求数列{an}的通项公式;(2)求数列{bn}的前n项和Tn.B提升题7.[2018山西八校第一次联考,11]已知数列{an}满足:a1=1,an+1=anan+2(n∈N*),若bn+1=(nλ)(1an+1),b1=λ,且数列{bn}是递增数列,则实数A.(2,+∞) B.(3,+∞) C.(∞,2) D.(∞,3)8.[2017湖北武汉四月调考,7]已知数列{an}满足a1=1,a2=13,若an(an1+2an+1)=3an1an+1(n≥2,n∈N*),则数列{an}的通项公式an=()A.12n-1 B.12n9.[2017辽宁省部分重点高中第三次联考,11]已知Sn为数列{an}的前n项和,且2am=am1+am+1(m∈N*,m≥2),若(a22)5+2016(a22)3+2017(a22)=2017,(a20162)5+2016(a20162)3+2017(a20162)=2017,则下列四个命题中真命题的序号为()①S2016=4032;②S2017=4034;③S2016<S2;④a2016a2<0.A.①② B.②③C.②④ D.①④10.[2018石家庄市重点高中高三摸底考试,15]已知数列{an}的前n项和为Sn,若a1=1,an≠0,anan+1=2Sn1,则a2n=.

11.[2018广东七校联考,17]已知{an}是递增数列,其前n项和为Sn,a1>1,且10Sn=(2an+1)(an+2)(n∈N*).(1)求数列{an}的通项公式an;(2)是否存在m,n,k∈N*,使得2(am+an)=ak成立?若存在,写出一组符合条件的m,n,k的值;若不存在,请说明理由.答案1.1121由a1+a2=4,a2=2a1+1,得a1=1.由an+1=Sn+1Sn=2Sn+1,得Sn+1=3Sn+1,所以Sn+1+12=3(Sn+12),所以{Sn+12}是以32为首项,3为公比的等比数列,所以Sn+12=2.2011由a1=1,且an+1an=n+1(n∈N*)得,an=a1+(a2a1)+(a3a2)+…+(anan1)=1+23+…+n=n(n+1)2,则1an=2n(n+1)=2(1n1n+1),故数列{1an}前10项的和2(1111)=203.12将a8=2代入an+1=11-an,可求得a7=12;再将a7=12代入an+1=11-an,可求得a6=1;再将a6=1代入an+1=11-an,可求得a5=24.(2)n1当n=1时,由已知Sn=23an+13,得S1=a1=23a1+13,即a1=1;当n≥2时,由已知得Sn1=23an1+13,所以an=SnSn1=(23an+13)(23an1+13)=23an23an1,所以an=2an1(n≥2),所以数列{5.(Ⅰ)由题意可得a2=12,a3=1(Ⅱ)由an2(2an+11)an2an+1=0,得2an+1(an+1)=an(an+因为{an}的各项都为正数,所以an+1a故{an}是首项为1,公比为12的等比数列,因此an=16.(Ⅰ)由an2+2an=4Sn+3①,可知an+12+2an+1=4S由②①可得an+12an2+2(an+1an)=2(an+1+an)=(an+1+an)(an+1an).因为an>0,所以an+1an=2.又a12+2a1=4a1+3,解得a1=1(舍去)或a1所以{an}是首项为3,公差为2的等差数列,其通项公式an=2n+1.(Ⅱ)由an=2n+1可知bn=1anan+1=1(2n设数列{bn}的前n项和为Tn,则Tn=b1+b2+…+bn=12[(1315)+(1517)+…=n3(2A基础题1.A依题意得1an+1=an+33an=1an+13,1an+11an=13,数列{1an}是以1a1=132.C解法一因为an=12an+1,a3+a4=2,所以an≠0,可得an+1an=2,所以{an}为等比数列,由an=amqnm,得a3+a3×243=2,解得a3=23,由此可得a4=a3×2=43,a5=a3×22=83,所以a4+a5=解法二已知an=12an+1,可得an+1=2an,所以a4+a5=2a3+2a4=2(a3+a4)=2×2=43.B由2,Sn,an成等差数列,得2Sn=an+2①,2Sn+1=an+1+2②,②①,得an+1an=1,又2a1=a1+2,所以a1=2,所以数列{an}是首项为2,公比为1的等比数列,所以S17=4.n·2n1an+12an=2n两边同除以2n+1,可得an+12n+1an2n=12,又a12=12,∴数列{an2n}是以12为首项,12为公差的等差数列,∴an25.(2)n1当n=1时,3S12a1=3a12a1=1,得a1=1;当n≥2时,3Sn-2an=1,3Sn-1-2an-1=1,6.(1)∵Sn=2n+12,∴当n=1时,a1=S1=21+12=2;当n≥2时,an=SnSn1=2n+12n=2n.又a1=2=21,∴an=2n.(2)由(1)知,bn=anSn=2·4n2n+1,∴Tn=b1+b2+b3+…+bn=2(41+42+43+…+4n)(22+23+…+2n+1)=2×4(1-4n)1-44(1B提升题7.C由an+1=anan+2,知1an+1=2an+1,即1an+1+1=2(1an+1),所以数列{1an+1}是首项为1a1+1=2,公比为2的等比数列,所以1an+1=2n,所以bn+1=(nλ)·2n,因为数列{bn}是递增数列,所以bn+1bn=(nλ)2n(n1λ)2n1=(8.B解法一an(an1+2an+1)=3an1an+1⇒1an+1+2an-1=3a又1a21a1=2,∴{1an+11an}是首项为2,公比为2的等比数列,则1an+11an=2n,即1an1a1=(1a21a1)+(1a31a解法二由a2(a1+2a3)=3a1a3,得a3=17,即可排除选项A,C,D.选B9.C构造函数f(x)=x5+2016x3+2017x,∵f(x)为奇函数且单调递增,依题意有f(a22)=2017,f(a20162)=2017,∴(a22)+(a20162)=0,∴a2+a2016=4.又2am=am1+am+1(m∈N*,m≥2),∴数列{an}为等差数列,且公差d≠0,∴a1+a2017=a2+a2016=4,则S2017=2017(a∵公差d≠0,故a2016≠a2017,S2016=2016(由题意知a2>2,a2016<2,∴d<0,S2016=S2017a2017=4034(4a1)=4030+a1,S2=a1+a2,若S2016<S2,则a2>4030,而此时(a22)5+2016(a22)3+2017(a22)=2017不成立,③错误;∵a2>2,a2016<2,∴a2016a2<0,④正确.故选C.10.2n+1因为a1=1,anan+1=2Sn1,所以a2=3,当n≥2时,anan+1an1an=2an,又an≠0,所以an+1an1=2,所以数列{a2n}是以3为首项,2为公差的等差数列,所以a2n=3+(n1)×2=2n+1.11.(1)由10a1=(2a1+1)(a1+2),得2a125a1+2=0,解得a1=2或a1=12.又a1>1,所以a1因为10Sn=(2an+1)(an+2),所以10Sn=2an2+5an+故10an+1=10Sn+110Sn=2an+12+5an+1+22an整理,得2(an+12an2)5(an+1即(an+1+