（天津?qū)Ｓ茫?020版高考數(shù)學(xué)大一輪復(fù)習(xí) 6.1 數(shù)列的概念及其表示精練.docx

6.1數(shù)列的概念及其表示【真題典例】挖命題【考情探究】考點(diǎn)內(nèi)容解讀5年考情預(yù)測(cè)熱度考題示例考向關(guān)聯(lián)考點(diǎn)數(shù)列的有關(guān)概念及性質(zhì)1.了解數(shù)列的概念,數(shù)列的通項(xiàng)公式2.了解數(shù)列是自變量為正整數(shù)的一類函數(shù),會(huì)用賦值法求數(shù)列的項(xiàng)2011天津,20,14分賦值法求數(shù)列的項(xiàng)、數(shù)列的通項(xiàng)公式不等式的證明分析解讀了解數(shù)列的概念和有關(guān)的表示方法,了解數(shù)列的通項(xiàng)公式、遞推公式,了解數(shù)列的通項(xiàng)公式與前n項(xiàng)和公式之間的關(guān)系,了解數(shù)列是自變量為正整數(shù)的一類函數(shù).考查數(shù)列的有關(guān)概念和性質(zhì),培養(yǎng)學(xué)生的創(chuàng)新能力、抽象概括能力.本節(jié)內(nèi)容在高考中分值約為5分,屬于中低檔題.破考點(diǎn)【考點(diǎn)集訓(xùn)】考點(diǎn)數(shù)列的有關(guān)概念及性質(zhì)1.在數(shù)列an中,a1=0,an+1=3+an1-3an,則a2016=()A.23B.3C.0D.-3答案D2.已知數(shù)列an滿足a1=1,且an=n(an+1-an)(nN*),則a2=;an=.答案2;n3.已知數(shù)列an滿足an=3an-1+3n-1(nN*,n2),且a1=5,則an=.答案(n+4)3n-14.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=an+1-2n+2,a2=2,則an=.答案2,n=12n-2,n>1煉技法【方法集訓(xùn)】方法1利用an與Sn的關(guān)系求通項(xiàng)1.已知數(shù)列an的前n項(xiàng)和為Sn,若3Sn=2an-3n,則a2018=()A.22018-1B.32018-6C.122018-72D.132018-103答案A2.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=2an-1,則S6a6=()A.6332B.3116C.12364D.127128答案A3.已知數(shù)列an的前n項(xiàng)和為Sn,且a1=1,Sn=(n+1)an2,則a2017=()A.2016B.2017C.4032D.4034答案B4.已知Sn為數(shù)列an的前n項(xiàng)和,且log2(Sn+1)=n+1,則數(shù)列an的通項(xiàng)公式為.答案an=3,n=12n,n2方法2利用遞推關(guān)系求數(shù)列的通項(xiàng)5.已知數(shù)列an中,a1=1,an+1=2an+1(nN*),Sn為其前n項(xiàng)和,則S5的值為()A.57B.61C.62D.63答案A6.在數(shù)列an中,a1=1,an+1=2anan+2,則數(shù)列an的通項(xiàng)an=.答案2n+17.已知數(shù)列an的前n項(xiàng)之和為Sn,若a1=2,an+1=an+2n-1+1,則S10=.答案1078過(guò)專題【五年高考】A組自主命題天津卷題組(2011天津,20,14分)已知數(shù)列an與bn滿足bnan+an+1+bn+1an+2=0,bn=3+(-1)n2,nN*,且a1=2,a2=4.(1)求a3,a4,a5的值;(2)設(shè)cn=a2n-1+a2n+1,nN*,證明cn是等比數(shù)列;(3)設(shè)Sk=a2+a4+a2k,kN*,證明k=14nSkak<76(nN*).解析(1)由bn=3+(-1)n2,nN*,可得bn=1,n為奇數(shù),2,n為偶數(shù).又bnan+an+1+bn+1an+2=0,當(dāng)n=1時(shí),a1+a2+2a3=0,由a1=2,a2=4,可得a3=-3;當(dāng)n=2時(shí),2a2+a3+a4=0,可得a4=-5;當(dāng)n=3時(shí),a3+a4+2a5=0,可得a5=4.(2)證明:對(duì)任意nN*,a2n-1+a2n+2a2n+1=0,2a2n+a2n+1+a2n+2=0,a2n+1+a2n+2+2a2n+3=0,-,得a2n=a2n+3,將代入,可得a2n+1+a2n+3=-(a2n-1+a2n+1),即cn+1=-cn(nN*).又c1=a1+a3=-1,故cn0,因此cn+1cn=-1.所以cn是等比數(shù)列.(3)證明:由(2)可得a2k-1+a2k+1=(-1)k,于是,對(duì)任意kN*且k2,有a1+a3=-1,-(a3+a5)=-1,a5+a7=-1,(-1)k(a2k-3+a2k-1)=-1.將以上各式相加,得a1+(-1)ka2k-1=-(k-1),即a2k-1=(-1)k+1(k+1),此式當(dāng)k=1時(shí)也成立.由式得a2k=(-1)k+1(k+3).從而S2k=(a2+a4)+(a6+a8)+(a4k-2+a4k)=-k,S2k-1=S2k-a4k=k+3,所以,對(duì)任意nN*,n2,k=14nSkak=m=1nS4m-3a4m-3+S4m-2a4m-2+S4m-1a4m-1+S4ma4m=m=1n2m+22m-2m-12m+2-2m+32m+1+2m2m+3=m=1n22m(2m+1)+3(2m+2)(2m-3)=223+m=2n52m(2m+1)+3(2n+2)(2n+3)<13+m=2n5(2m-1)(2m+1)+3(2n+2)(2n+3)=13+5213-15+15-17+12n-1-12n+1+3(2n+2)(2n+3)=13+56-5212n+1+3(2n+2)(2n+3)<76.對(duì)于n=1,不等式顯然成立.思路分析本題主要考查等比數(shù)列的定義、數(shù)列求和的基礎(chǔ)知識(shí)和基本計(jì)算.(1)由已知條件bn=3+(-1)n2,bnan+an+1+bn+1an+2=0,a1=2,a2=4,依次代入n=1,2,3,求出a3,a4,a5的值.(2)由bn=1,n為奇數(shù),2,n為偶數(shù)和bnan+an+1+bn+1an+2=0得出a2n-1,a2n,a2n+1,a2n+2,a2n+3間的關(guān)系式,此步的目的是與cn=a2n-1+a2n+1形式統(tǒng)一,從而導(dǎo)出cn+1,cn的關(guān)系式,進(jìn)而證明cn是等比數(shù)列.(3)由(2)問(wèn)有a2k-1+a2k+1=(-1)k,通過(guò)累加得a2k-1=(-1)k+1(k+1),則有a2k=(-1)k+1(k+3).通過(guò)a2k,a2k-1的通項(xiàng)求出S2k-1,S2k的通項(xiàng),代入到k=14nSkak,通過(guò)放縮推導(dǎo)證明.B組統(tǒng)一命題、省(區(qū)、市)卷題組1.(2018課標(biāo),14,5分)記Sn為數(shù)列an的前n項(xiàng)和.若Sn=2an+1,則S6=.答案-632.(2015課標(biāo),16,5分)設(shè)Sn是數(shù)列an的前n項(xiàng)和,且a1=-1,an+1=SnSn+1,則Sn=.答案-1n3.(2016浙江,13,6分)設(shè)數(shù)列an的前n項(xiàng)和為Sn.若S2=4,an+1=2Sn+1,nN*,則a1=,S5=.答案1;121C組教師專用題組1.(2013課標(biāo),14,5分)若數(shù)列an的前n項(xiàng)和Sn=23an+13,則an的通項(xiàng)公式是an=.答案(-2)n-12.(2013安徽,14,5分)如圖,互不相同的點(diǎn)A1,A2,An,和B1,B2,Bn,分別在角O的兩條邊上,所有AnBn相互平行,且所有梯形AnBnBn+1An+1的面積均相等.設(shè)OAn=an.若a1=1,a2=2,則數(shù)列an的通項(xiàng)公式是.答案an=3n-23.(2016課標(biāo),17,12分)已知各項(xiàng)都為正數(shù)的數(shù)列an滿足a1=1,an2-(2an+1-1)an-2an+1=0.(1)求a2,a3;(2)求an的通項(xiàng)公式.解析(1)由題意得a2=12,a3=14.(5分)(2)由an2-(2an+1-1)an-2an+1=0得2an+1(an+1)=an(an+1).因?yàn)閍n的各項(xiàng)都為正數(shù),所以an+1an=12.故an是首項(xiàng)為1,公比為12的等比數(shù)列,因此an=12n-1.(12分)評(píng)析本題主要考查了數(shù)列的遞推公式及等比數(shù)列的定義,屬基礎(chǔ)題.4.(2014大綱全國(guó),17,10分)數(shù)列an滿足a1=1,a2=2,an+2=2an+1-an+2.(1)設(shè)bn=an+1-an,證明bn是等差數(shù)列;(2)求an的通項(xiàng)公式.解析(1)證明:由an+2=2an+1-an+2得,an+2-an+1=an+1-an+2,即bn+1=bn+2.又b1=a2-a1=1.所以bn是首項(xiàng)為1,公差為2的等差數(shù)列.(2)由(1)得bn=1+2(n-1),即an+1-an=2n-1.于是k=1n(ak+1-ak)=k=1n(2k-1),所以an+1-a1=n2,即an+1=n2+a1.又a1=1,所以an的通項(xiàng)公式為an=n2-2n+2.【三年模擬】一、選擇題(每小題5分,共20分)1.(2018天津南開(kāi)基礎(chǔ)訓(xùn)練,5)在數(shù)列an中,a1=3,an+an-1=4(n2),則a2018=()A.3B.1C.-3D.4答案B2.(2017天津一中3月月考,6)數(shù)列an滿足a1=1,且對(duì)任意的nN*都有an+1=a1+an+n,則1a1+1a2+1a2016=()A.20152016B.20162017C.40322017D.40342017答案C3.(2017天津河?xùn)|二模,7)若數(shù)列an,bn的通項(xiàng)公式分別為an=(-1)n+2016a,bn=2+(-1)n+2017n,且對(duì)任意nN*,an<bn恒成立,則實(shí)數(shù)a的取值范圍是()A.-1,12B.-1,1)C.-2,1)D.-2,32答案D4.(2019屆天津耀華中學(xué)統(tǒng)練(2),9)設(shè)a1=2,an+1=2an+1,bn=an+2an-1,nN*,則數(shù)列bn的通項(xiàng)公式為()A.bn=2n+2B.bn=4nC.bn=2n+1D.bn=4n答案C二、填空題(每小題5分,共30分)5.(2019屆天津七校聯(lián)考,12)已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=32an-12,則log3a100=.答案996.(2019屆天津新華中學(xué)期中,11)若數(shù)列an滿足a1=2,an+1=an+2n+n,則數(shù)列an的通項(xiàng)公式為an=.答案2n+n(n-1)27.(2019屆天津耀華中學(xué)第二次月考,11)設(shè)數(shù)列an的前n項(xiàng)和為Sn,若Sn=2an+n(nN*),則數(shù)列an的通項(xiàng)公式為an=.答案1-2n8.(2019屆天津耀華中學(xué)統(tǒng)練(2),13)數(shù)列an的前n項(xiàng)和為Sn=n2+n+1,bn=(-1)nan(nN*),則數(shù)列bn的前50項(xiàng)的和為.答案499.(2019屆天津一中1月月考,10)已知數(shù)列an的前n項(xiàng)和為Sn,且a2=4,S4=30,n2時(shí),an+1+an-1=2(an+1),則an的通項(xiàng)公式為an=.答案n210.(2019屆天津耀華中學(xué)統(tǒng)練(2),15)已知數(shù)列an,a1=1,nN*,an+1=an+2n+1n2(n+1)2,則an的通項(xiàng)公式為.答案an=2-1n2三、解答題(共35分)11.(2017天津和平期末,18)設(shè)數(shù)列an滿足a1=1,an+1=an+32n-1.(1)求數(shù)列an的通項(xiàng)公式;(2)若bnan=n,求數(shù)列bn的前n項(xiàng)和Sn.解析(1)a1=1,an+1-an=32n-1,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+320+321+32n-2=1+3(20+21+2n-2)=1+31(1-2n-1)1-2=32n-1-2(n2),當(dāng)n=1時(shí),321-1-2=1=a1成立,數(shù)列an的通項(xiàng)公式為an=32n-1-2(nN*).(2)bn=nan=3n2n-1-2n,b1=3120-2,b2=3221-4,b3=3322-6,bn=3n2n-1-2n,Sn=b1+b2+b3+bn=3(120+221+322+n2n-1)-(2+4+6+2n).設(shè)Tn=120+221+322+n2n-1,則2Tn=121+222+(n-1)2n-1+n2n,-,得-Tn=(20+21+22+2n-1)-n2n=(2n-1)-n2n,Tn=(n-1)2n+1,Sn=3(n-1)2n+3-2(1+2+3+n)=3(n-1)2n-n(n+1)+3.12.(2018天津薊州一中模擬,18)在數(shù)列an中,an>0,其前n項(xiàng)和Sn滿足Sn2-(n2+2n-1)Sn-(n2+2n)=0.(1)求an的通項(xiàng)公式;(2)若bn=an-52n,求b2+b4+b2n.解析(1)由Sn2-(n2+2n-1)Sn-(n2+2n)=0,得Sn-(n2+2n)(Sn+1)=0,由an>0,可知Sn>0,故Sn=n2+2n.當(dāng)n2時(shí),an=Sn-Sn-1=(n2+2n)-(n-1)2+2(n-1)=2n+1;當(dāng)n=1時(shí),a1=S1=3,符合上式,則數(shù)列an的通項(xiàng)公式為an=2n+1(nN*).(2)依題意,得bn=2n-42n=n-22n-1,則b2n=2n-222n-1=(n-1)14n-1,nN*,設(shè)Tn=b2+b4+b2n,故Tn=0+14+242+343+n-14n-1,而4Tn=1+24+342+n-14n-2.-得3Tn=1+14+142+14n-2-n-14n-1=1-14n-11-14-n-14n-1=134-3n+14n-1,故Tn=194-3n+14n-1.思路分析本題主要考查數(shù)列的通項(xiàng)公式和前n項(xiàng)和的求法,解題時(shí)要認(rèn)真審題,注意錯(cuò)位相減法的合理運(yùn)用.(1)由Sn2-(n2+2n-1)Sn-(n2+2n)=0,得Sn=n2+2n,再由an=Sn-Sn-1,能求出數(shù)列an的通項(xiàng)公式;(2)由(1)知bn=2n-42n=n-22n-1,利用錯(cuò)位相減法求出Tn.13.(2018天津河北一模,18)已知數(shù)列an的前n項(xiàng)和為Sn,且an=2-2Sn(nN*),數(shù)列bn是等差數(shù)列,且b5=14,b7=20.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)若cn=anbn,nN*,求數(shù)列cn的前n項(xiàng)和Tn.解析(1)an=2-2Sn(nN*),an-1=2-2Sn-1(n2),an-an-1=-2an,即an=13an-1,當(dāng)n=1時(shí),a1=2-2S1,解得a1=23,數(shù)列an是以23為首項(xiàng),13為公比的等比數(shù)列,an=213n.設(shè)數(shù)列bn的公差為d,則b1+4d=14,b1+6d=20,解得b1=2,d=3,bn=3n-1.(2)cn=anbn=2(3n-1)13n,Tn=2213+5132+8133+(3n-1)13n,13Tn=22132+5133+8134+(3n-1)13n+1,兩式相減可得,23Tn=223+3132+133+134+13n-(3n-1)13n+1=223+3191-13n-11-13-(3n-1)13n+1,化簡(jiǎn)可得Tn=72-6n+723n.