溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)節(jié)合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。課時(shí)分層提升練三十?dāng)?shù)列的概念與簡(jiǎn)單表示法……25分鐘50分一、選擇題(每小題5分,共35分)1.數(shù)列1,12,13,14,1A.(-1)C.(-1)n【解析】選A.設(shè)此數(shù)列的通項(xiàng)公式為an,因?yàn)槠鏀?shù)項(xiàng)為負(fù)數(shù),偶數(shù)項(xiàng)為正數(shù),每一項(xiàng)的絕對(duì)值為1n故其通項(xiàng)公式為an=(-12.已知數(shù)列{an}的通項(xiàng)公式是an=23n,則該數(shù)列的第五項(xiàng)是 ()【解析】選A.因?yàn)閿?shù)列{an}的通項(xiàng)公式是an=23n,所以a5=23×5=13.3.數(shù)列23,45,67,89,…1617 1819 20【解析】選n=(1)n+12n2n+1,a4.已知數(shù)列{an}對(duì)于任意p,q∈N*,有ap+aq=ap+q,若a1=19,則a36A.136 B.1【解析】選D.因?yàn)閍p+q=ap+aq,所以a36=a32+a4=2a16+a4=4a8+a4=8a4+a4=18a2=36a1=4.5.數(shù)列{an}中,a1=1,對(duì)于所有的n≥2,n∈N*都有a1·a2·a3·…·an=n2,則a3+a5= ()A.6116 B.259 C.2516【解析】選A.方法一:令n=2,3,4,5,分別求出a3=94,a5=25所以a3+a5=6116方法二:當(dāng)n≥2時(shí),a1·a2·a3·…·an=n2.當(dāng)n≥3時(shí),a1·a2·a3·…·an1=(n1)2.兩式相除得an=nn-12,所以a3=94所以a3+a5=61166.已知數(shù)列{an}滿足a1=60,an+1an=2n,則annA.292 D.【解析】選A.因?yàn)閍n+1an=2n,所以當(dāng)n≥2時(shí),an=a1+(a2a1)+(a3a2)+…+(anan1)=60+2+4+…+2(n1)=n(n1)+60=n2n+60,所以ann=n2-n+60n=n+60由函數(shù)性質(zhì)可知,f(x)在區(qū)間[2,215)上單調(diào)遞減,在區(qū)間(215,+∞)上單調(diào)遞增,又7<215<8,n為正整數(shù),故當(dāng)n=7時(shí),ann=7+607當(dāng)n=8時(shí),ann=8+608且292<1027<所以ann的最小值為7.數(shù)列{an}的前n項(xiàng)和為Sn,若a1=1,an+1=3Sn(n≥1),則a6= ()A.3×44 B.3×44+15 5+1【解析】選A.方法一:a1=1,a2=3S1=3,a3=3S2=12=3×41,a4=3S3=48=3×42,a5=3S4=3×43,a6=3S5=3×44.方法二:當(dāng)n≥1時(shí),an+1=3Sn,則an+2=3Sn+1,所以an+2an+1=3Sn+13Sn=3an+1,即an+2=4an+1,所以該數(shù)列從第2項(xiàng)開始是以4為公比的等比數(shù)列,又a2=3S1=3a1=3,所以an=1(n=1),3×4n-二、填空題(每小題5分,共15分)8.已知非零數(shù)列{an}的遞推公式為an=nn-1·an-1(n>1),且a1=1,【解析】依次對(duì)遞推公式中的n賦值,當(dāng)n=2時(shí),a2=2a1;當(dāng)n=3時(shí),a3=32a2=3a1;當(dāng)n=4時(shí),a4=43a3=4a答案:49.若數(shù)列{an}為1,2,2,3,3,4,4,…,則該數(shù)列的通項(xiàng)公式an=________.

【解析】數(shù)列{an}為1,2,2,3,3,4,4,…,可得奇數(shù)項(xiàng)分別為:1,2,3,4,…,可得an=n+1偶數(shù)項(xiàng)分別為2,3,4,…,可得an=n+2則該數(shù)列的通項(xiàng)公式an=n答案:n10.若數(shù)列{an}的前n項(xiàng)和Sn=23an+13,則{an}的通項(xiàng)公式是an=【解析】當(dāng)n=1時(shí),S1=23a1+13,所以a當(dāng)n≥2時(shí),an=SnSn1=23an+1323an-1所以an=2an1,即an所以{an}是以1為首項(xiàng)的等比數(shù)列,其公比為2,所以an=1×(2)n1,即an=(2)n1.答案:(2)n1……15分鐘30分1.(5分)已知數(shù)列11,21,12,31,22,13,41,32,23,14,…,依它的前10項(xiàng)的規(guī)律,這個(gè)數(shù)列的第A.131 B.163 C.64 【解析】選D.觀察數(shù)列:11,21,12,31,22,13,41,32,23,14,…,得出:并且在每一個(gè)k段內(nèi),是k個(gè)分?jǐn)?shù)(k∈N*,k≥3),且它們的分子分母和為k+1(k∈N*,k≥3);由k=63時(shí),k(k+1)2故a2018在第64段中,所以該數(shù)列的第2018項(xiàng)a2018為第64組的第2項(xiàng),故a2018=6322.(5分)在數(shù)列{an}中,a1=2,an+1=an+ln1+1n,則aA.2+lnn B.2+(n1)lnnC.2+nlnn D.1+n+lnn【解析】n+1an=lnn+1n=ln(n+1)lnn,所以a2a1=ln2a3a2=ln3ln2,a4a3=ln4ln3,anan-1=lnn把以上各式分別相加,得ana1=lnnln1,即an=2+lnn.巧思妙解:由an+1an=ln(n+1)lan=(anan-1)+(an-1an-=lnnln(n1)+ln(n1)ln(n2)+…+ln2ln1+2=lnn+2.3.(5分)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 ()n=2n n=3n=2n1 n=2n+1【解析】log2(Sn+1)=n+1,得Sn+1=2n+1.當(dāng)n=1時(shí),a1=S1=3;當(dāng)n≥2時(shí),an=SnSn1=2n.所以數(shù)列{an}的通項(xiàng)公式為an=34.(5分)在數(shù)列{an}中,若a4=1,a12=5,且任意連續(xù)三項(xiàng)的和都是15,則a2020=________.

【解析】因?yàn)槿我膺B續(xù)三項(xiàng)的和都是15,所以an+an+1+an+2=15,同時(shí)an+1+an+2+an+3=15,則an+an+1+an+2=an+1+an+2+an+3,即an+3=an,即數(shù)列是周期為3的周期數(shù)列,則由a4=1,a12=5,得a4=a1=1,a12=a9=a6=a3=5,則由a1+a2+a3=15,得a2=9,則a2020=a673×3+1=a1=1.答案:15.(10分)已知數(shù)列{an}的各項(xiàng)均不為0,其前n項(xiàng)和為Sn,且a1=1,2Sn=anan+1,求Sn的表達(dá)式.【解析】當(dāng)n=1時(shí),2S1=a1a2,即2a1=a1a2,所以a2=2.當(dāng)n≥2時(shí),2Sn=anan+1