文科數(shù)學(xué)一輪復(fù)習(xí)高考幫全國版試題第6章第1講數(shù)列（考題幫數(shù)學(xué)文）

第一講數(shù)列的概念與簡單表示法題組數(shù)列的通項公式及前n項和1.[2016浙江,13,6分]設(shè)數(shù)列{an}的前n項和為Sn.若S2=4,an+1=2Sn+1,n∈N*,則a1=,S5=.

2.[2015江蘇,11,5分][文]設(shè)數(shù)列{an}滿足a1=1,且an+1an=n+1(n∈N*),則數(shù)列{1an}前10項的和為3.[2014新課標(biāo)全國Ⅱ,16,5分][文]數(shù)列{an}滿足an+1=11-an,a8=2,則a14.[2013新課標(biāo)全國Ⅰ,14,5分]若數(shù)列{an}的前n項和Sn=23an+13,則{an}的通項公式是an=5.[2016全國卷Ⅲ,17,12分][文]已知各項都為正數(shù)的數(shù)列{an}滿a1=1,an2(2an+11)an2an+1=(Ⅰ)求a2,a3;(Ⅱ)求{an}的通項公式.6.[2015新課標(biāo)全國Ⅰ,17,12分]Sn為數(shù)列{an}的前n項和,已知an>0,an2+2an=4Sn+(Ⅰ)求{an}的通項公式;(Ⅱ)設(shè)bn=1anan+1,求數(shù)列{bA組基礎(chǔ)題1.[2017云南省11校調(diào)考,3]在數(shù)列{an}中,a1=3,an+1=3anan+3,則a4A.34 B.1 C.43 2.[2017貴州省高招適應(yīng)性考試,3]已知數(shù)列{an}滿足an=12an+1,若a3+a4=2,則a4+a5=()A.12 B.1 C.4 3.[2017昆明市高三質(zhì)檢,5]已知數(shù)列{an}的前n項和為Sn,且2,Sn,an成等差數(shù)列,則S17=()A.0 B.2 C.2 D.344.[2018惠州市二調(diào),15]已知數(shù)列{an}滿足a1=1,an+12an=2n(n∈N*),則數(shù)列{an}的通項公式an=.

5.[2017鄭州市第三次質(zhì)量預(yù)測,14]若數(shù)列{an}的前n項和為Sn,且3Sn2an=1,則{an}的通項公式是an=.

6.[2018南昌市摸底調(diào)研,17]已知數(shù)列{an}的前n項和Sn=2n+12,記bn=anSn(n∈N*).(1)求數(shù)列{an}的通項公式;(2)求數(shù)列{bn}的前n項和Tn.B提升題7.[2018山西八校第一次聯(lián)考,11]已知數(shù)列{an}滿足:a1=1,an+1=anan+2(n∈N*),若bn+1=(nλ)(1an+1),b1=λ,且數(shù)列{bn}是遞增數(shù)列,則實數(shù)A.(2,+∞) B.(3,+∞) C.(∞,2) D.(∞,3)8.[2017湖北武漢四月調(diào)考,7]已知數(shù)列{an}滿足a1=1,a2=13,若an(an1+2an+1)=3an1an+1(n≥2,n∈N*),則數(shù)列{an}的通項公式an=()A.12n-1 B.12n9.[2017遼寧省部分重點高中第三次聯(lián)考,11]已知Sn為數(shù)列{an}的前n項和,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,則下列四個命題中真命題的序號為()①S2016=4032;②S2017=4034;③S2016<S2;④a2016a2<0.A.①② B.②③C.②④ D.①④10.[2018石家莊市重點高中高三摸底考試,15]已知數(shù)列{an}的前n項和為Sn,若a1=1,an≠0,anan+1=2Sn1,則a2n=.

11.[2018廣東七校聯(lián)考,17]已知{an}是遞增數(shù)列,其前n項和為Sn,a1>1,且10Sn=(2an+1)(an+2)(n∈N*).(1)求數(shù)列{an}的通項公式an;(2)是否存在m,n,k∈N*,使得2(am+an)=ak成立?若存在,寫出一組符合條件的m,n,k的值;若不存在,請說明理由.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為首項,3為公比的等比數(shù)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,則1an=2n(n+1)=2(1n1n+1),故數(shù)列{1an}前10項的和2(1111)=203.12將a8=2代入an+1=11-an,可求得a7=12;再將a7=12代入an+1=11-an,可求得a6=1;再將a6=1代入an+1=11-an,可求得a5=24.(2)n1當(dāng)n=1時,由已知Sn=23an+13,得S1=a1=23a1+13,即a1=1;當(dāng)n≥2時,由已知得Sn1=23an1+13,所以an=SnSn1=(23an+13)(23an1+13)=23an23an1,所以an=2an1(n≥2),所以數(shù)列{5.(Ⅰ)由題意可得a2=12,a3=1(Ⅱ)由an2(2an+11)an2an+1=0,得2an+1(an+1)=an(an+因為{an}的各項都為正數(shù),所以an+1a故{an}是首項為1,公比為12的等比數(shù)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).因為an>0,所以an+1an=2.又a12+2a1=4a1+3,解得a1=1(舍去)或a1所以{an}是首項為3,公差為2的等差數(shù)列,其通項公式an=2n+1.(Ⅱ)由an=2n+1可知bn=1anan+1=1(2n設(shè)數(shù)列{bn}的前n項和為Tn,則Tn=b1+b2+…+bn=12[(1315)+(1517)+…=n3(2A基礎(chǔ)題1.A依題意得1an+1=an+33an=1an+13,1an+11an=13,數(shù)列{1an}是以1a1=132.C解法一因為an=12an+1,a3+a4=2,所以an≠0,可得an+1an=2,所以{an}為等比數(shù)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成等差數(shù)列,得2Sn=an+2①,2Sn+1=an+1+2②,②①,得an+1an=1,又2a1=a1+2,所以a1=2,所以數(shù)列{an}是首項為2,公比為1的等比數(shù)列,所以S17=4.n·2n1an+12an=2n兩邊同除以2n+1,可得an+12n+1an2n=12,又a12=12,∴數(shù)列{an2n}是以12為首項,12為公差的等差數(shù)列,∴an25.(2)n1當(dāng)n=1時,3S12a1=3a12a1=1,得a1=1;當(dāng)n≥2時,3Sn-2an=1,3Sn-1-2an-1=1,6.(1)∵Sn=2n+12,∴當(dāng)n=1時,a1=S1=21+12=2;當(dāng)n≥2時,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提升題7.C由an+1=anan+2,知1an+1=2an+1,即1an+1+1=2(1an+1),所以數(shù)列{1an+1}是首項為1a1+1=2,公比為2的等比數(shù)列,所以1an+1=2n,所以bn+1=(nλ)·2n,因為數(shù)列{bn}是遞增數(shù)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}是首項為2,公比為2的等比數(shù)列,則1an+11an=2n,即1an1a1=(1a21a1)+(1a31a解法二由a2(a1+2a3)=3a1a3,得a3=17,即可排除選項A,C,D.選B9.C構(gòu)造函數(shù)f(x)=x5+2016x3+2017x,∵f(x)為奇函數(shù)且單調(diào)遞增,依題意有f(a22)=2017,f(a20162)=2017,∴(a22)+(a20162)=0,∴a2+a2016=4.又2am=am1+am+1(m∈N*,m≥2),∴數(shù)列{an}為等差數(shù)列,且公差d≠0,∴a1+a2017=a2+a2016=4,則S2017=2017(a∵公差d≠0,故a2016≠a2017,S2016=2016(由題意知a2>2,a2016<2,∴d<0,S2016=S2017a2017=4034(4a1)=4030+a1,S2=a1+a2,若S2016<S2,則a2>4030,而此時(a22)5+2016(a22)3+2017(a22)=2017不成立,③錯誤;∵a2>2,a2016<2,∴a2016a2<0,④正確.故選C.10.2n+1因為a1=1,anan+1=2Sn1,所以a2=3,當(dāng)n≥2時,anan+1an1an=2an,又an≠0,所以an+1an1=2,所以數(shù)列{a2n}是以3為首項,2為公差的等差數(shù)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因為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+