（福建專版）2019高考數(shù)學一輪復(fù)習 課時規(guī)范練27 數(shù)列的概念與表示 文.docx

課時規(guī)范練27數(shù)列的概念與表示基礎(chǔ)鞏固組1.數(shù)列1,23,35,47,59,的一個通項公式an=() A.n2n+1B.n2n-1C.n2n-3D.n2n+32.已知數(shù)列an的前n項和為Sn,且Sn=2(an-1),則a2等于()A.4B.2C.1D.-23.(2017江西上饒模擬)已知數(shù)列an滿足an+1+an=n,若a1=2,則a4-a2=()A.4B.3C.2D.14.已知數(shù)列an滿足a1=0,an+1=an+2n-1,則數(shù)列an的一個通項公式為()A.an=n-1B.an=(n-1)2C.an=(n-1)3D.an=(n-1)45.(2017吉林市模擬改編)若數(shù)列an滿足a1=12,an=1-1an-1(n2,且nN*),則a2 018等于()A.-1B.12C.1D.2導學號241907526.已知數(shù)列an的首項a1=1,其前n項和Sn=n2an(nN*),則a9=()A.136B.145C.155D.1667.(2017寧夏銀川二模,文16)已知數(shù)列an滿足a1=2,且a12+a23+a34+an-1n=an-2(n2),則an的通項公式為.8.已知數(shù)列an的通項公式為an=(n+2)78n,則當an取得最大值時,n=.9.已知各項都為正數(shù)的數(shù)列an滿足an+12-an+1an-2an2=0,且a1=2,則an=.10.(2017廣東江門一模,文17)已知正項數(shù)列an的前n項和為Sn,Sn=12an(an+1),nN*.(1)求數(shù)列an的通項公式;(2)若bn=1Sn,求數(shù)列bn的前n項和Tn.綜合提升組11.(2017河南鄭州、平頂山、濮陽二模,文8)已知數(shù)列an滿足an+1=an-an-1(n2),a1=m,a2=n,Sn為數(shù)列an的前n項和,則S2 017的值為()A.2 017n-mB.n-2 017mC.mD.n12.已知函數(shù)f(x)是定義在區(qū)間(0,+)內(nèi)的單調(diào)函數(shù),且對任意的正數(shù)x,y都有f(xy)=f(x)+f(y).若數(shù)列an的前n項和為Sn,且滿足f(Sn+2)-f(an)=f(3)(nN*),則an等于()A.2n-1B.nC.2n-1D.32n-113.(2017山西晉中二模)我們可以利用數(shù)列an的遞推公式an=n,n為奇數(shù),an2,n為偶數(shù)(nN*),求出這個數(shù)列各項的值,使得這個數(shù)列中的每一項都是奇數(shù),則a64+a65=.導學號2419075314.(2017山西呂梁二模)在數(shù)列an中,已知a2n=a2n-1+(-1)n,a2n+1=a2n+n,a1=1,則a20=.15.已知數(shù)列an的前n項和為Sn,Sn=2an-n,則an=.創(chuàng)新應(yīng)用組16.(2017河南洛陽一模,文7)意大利著名數(shù)學家斐波那契在研究兔子繁殖問題時,發(fā)現(xiàn)有這樣一列數(shù):1,1,2,3,5,8,13,.該數(shù)列的特點是:前兩個數(shù)都是1,從第三個數(shù)起,每一個數(shù)都等于它前面兩個數(shù)的和,人們把這樣的一列數(shù)所組成的數(shù)列an稱為“斐波那契數(shù)列”,則(a1a3-a22)(a2a4-a32)(a3a5-a42)(a2 015a2 017-a2 0162)=()A.1B.-1C.2 017D.-2 01717.已知數(shù)列an中,a1=-1,an+1=2an+3n-1(nN*),求數(shù)列an的通項公式.答案：1.B由已知得,數(shù)列可寫成11,23,35,故通項為n2n-1.2.A由Sn=2(an-1),得a1=2(a1-1),即a1=2,又a1+a2=2(a2-1),所以a2=4.3.D由an+1+an=n,得an+2+an+1=n+1,兩式相減得an+2-an=1,令n=2,得a4-a2=1.4.B因為a1=0,an+1=an+2n-1,所以a2=0+1=1,a3=1+3=4,a4=4+5=9,故數(shù)列an的一個通項公式為an=(n-1)2.5.Aa1=12,an=1-1an-1(n2,且nN*),a2=1-1a1=1-112=-1,a3=1-1a2=1-1-1=2,a4=1-1a3=1-12=12,依此類推,可得an+3=an,a2 018=a6723+2=a2=-1,故選A.6.B由Sn=n2an,得Sn+1=(n+1)2an+1,所以an+1=(n+1)2an+1-n2an,化簡得(n+2)an+1=nan,即an+1an=nn+2,所以a9=a9a8a8a7a2a1a1=810796824131=290=145.7.an=n+1a12+a23+a34+an-1n=an-2(n2),a12+a23+a34+ann+1=an+1-2(n2),-得ann+1=an+1-an,整理得an+1an=n+2n+1,an+1n+2ann+1=1,又a11+1=1,數(shù)列ann+1是以1為首項,1為公比的等比數(shù)列,即常數(shù)列1,an=n+1.8.5或6由題意令anan-1,anan+1,(n+2)78n(n+1)78n-1,(n+2)78n(n+3)78n+1,解得n6,n5.n=5或n=6.9.2nan+12-an+1an-2an2=0,(an+1+an)(an+1-2an)=0.數(shù)列an的各項均為正數(shù),an+1+an>0,an+1-2an=0,即an+1=2an(nN*),數(shù)列an是以2為公比的等比數(shù)列.a1=2,an=2n.10.解 (1)a1=S1=12a1(a1+1),a1>0,解得a1=1.nN*,an+1=Sn+1-Sn=12an+1(an+1+1)-12an(an+1),移項整理并因式分解得(an+1-an-1)(an+1+an)=0,因為an是正項數(shù)列,所以an+1+an>0,所以an+1-an-1=0,an+1-an=1.所以an是首項a1=1、公差為1的等差數(shù)列,所以an=n.(2)由(1)得Sn=12an(an+1)=12n(n+1),bn=1Sn=2n(n+1)=2n-2n+1,Tn=b1+b2+bn=21-22+22-23+2n-2n+1=21-2n+1=2nn+1.11.Can+1=an-an-1(n2),a1=m,a2=n,a3=n-m,a4=-m,a5=-n,a6=m-n,a7=m,a8=n,an+6=an.則S2 017=S3366+1=336(a1+a2+a6)+a1=3360+m=m.12.D由題意知f(Sn+2)=f(an)+f(3)=f(3an)(nN*),Sn+2=3an,Sn-1+2=3an-1(n2),兩式相減,得2an=3an-1(n2),則anan-1=32(n2).又n=1時,S1+2=3a1=a1+2,a1=1.數(shù)列an是首項為1,公比為32的等比數(shù)列.an=32n-1.13.66由題得,這個數(shù)列各項的值分別為1,1,3,1,5,3,7,1,9,5,11,3,a64+a65=a32+65=a16+65=a8+65=a4+65=1+65=66.14.46由a2n=a2n-1+(-1)n,得a2n-a2n-1=(-1)n,由a2n+1=a2n+n,得a2n+1-a2n=n,a2-a1=-1,a4-a3=1,a6-a5=-1,a20-a19=1,10個式子之和為0,a3-a2=1,a5-a4=2,a7-a6=3,a19-a18=9,9個式子之和為9(1+9)2=45.累加得a20-a1=45.又a1=1,故a20=46,故答案為46.15.2n-1當n2時,an=Sn-Sn-1=2an-n-2an-1+(n-1),即an=2an-1+1,an+1=2(an-1+1).又a1=S1=2a1-1,a1=1.數(shù)列an+1是以首項為a1+1=2,公比為2的等比數(shù)列,an+1=22n-1=2n,an=2n-1.16.Ba1a3-a22=12-12=1,a2a4-a32=13-22=-1,a3a5-a42=25-32=1,a2 015a2 017-a2 0162=1.(a1a3-a22)(a2a4-a32)(a3a5-a42)(a2 015a2 017-a2 0162)=11 008(-1)1 007=-1.17.解 an+1=2an+3n-1(nN*),a1=-1,a2=0.當n2時,an=2an-1+3n-4,由-可得an+1-an=2an-2an-1+3,即an+1-an+3=2(an-an-1+3),數(shù)列an-an-1+3為等比數(shù)列,首項為4,公比為2.an-an-1+3=42n-2,an-an-1=2n-3.an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2n-3+2n-1-3+22-3-1=4(2n-1-1)2-1-3(n-1)-1=2n+1-3n-2.