(福建專版)2019高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練27數(shù)列概念與表示文

課時(shí)規(guī)范練27數(shù)列的觀點(diǎn)與表示基礎(chǔ)穩(wěn)固組1.數(shù)列1,,的一個(gè)通項(xiàng)公式an=( )A.B.C.D.--2.已知數(shù)列{an}的前n項(xiàng)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}知足an+1+an=n,若a1=2,則a4-a2=()A.4B.3C.2D.14.已知數(shù)列{an}知足a1=0,an+1=an+2n-1,則數(shù)列{an}的一個(gè)通項(xiàng)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{a}知足a=,a=1-(n≥,且n∈N),則a2018等于( )n1n*-A.-1B.C.1D.2?導(dǎo)學(xué)號24190752?6.已知數(shù)列{an}的首項(xiàng)a1=1,其前n項(xiàng)和Sn=n2an(n∈N*),則a9=()A.B.C.D.7.(2017

寧夏銀川二模

,文

16)已知數(shù)列

{an}知足

a1=2,且

++-

=an-2(n≥

),

則{an}的通項(xiàng)公式為.8.已知數(shù)列{an}的通項(xiàng)公式為an=(n+2),則當(dāng)an獲得最大值時(shí),n=.9.已知各項(xiàng)都為正數(shù)的數(shù)列{n}知足-an+1n20,且12,則n.aa-=a=a=10.(2017廣東江門一模,文17)已知正項(xiàng)數(shù)列{an}的前n項(xiàng)和為Sn,Sn=an(an+1),*n∈N.求數(shù)列{an}的通項(xiàng)公式;若bn=,求數(shù)列{bn}的前n項(xiàng)和Tn.綜合提高組11.(2017河南鄭州、平頂山、濮陽二模,文8)已知數(shù)列{an}知足an+1=an-an-1(n≥),a1=m,a2=n,Sn為數(shù)列{an}的前n項(xiàng)和,則S2017的值為()A.2017n-mB.n-2017mC.mD.n12.已知函數(shù)f(x)是定義在區(qū)間(0,+∞)內(nèi)的單一函數(shù),且對隨意的正數(shù)x,y都有f(xy)=f(x)+f(y).若數(shù)列{an}的前n項(xiàng)和為Sn,且知足f(Sn+2)-f(an)=f(3)(n∈N*),則an等于()n-1B.nC.2n-1D.-A.213.(2017山西晉中二模)我們能夠利用數(shù)列,為奇數(shù),*aa=nnn,為偶數(shù)項(xiàng)的值,使得這個(gè)數(shù)列中的每一項(xiàng)都是奇數(shù),則a64+a65=.?導(dǎo)學(xué)號24190753?14.(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(xiàng)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)意大利有名數(shù)學(xué)家斐波那契在研究兔子生殖問題時(shí),發(fā)現(xiàn)有這樣一列數(shù):,,,,,,,.該數(shù)列的特色是:前兩個(gè)數(shù)都是1,從第三個(gè)數(shù)起,每一個(gè)數(shù)都等于它前面兩個(gè)數(shù)的和,人們把這樣的一列數(shù)所構(gòu)成的數(shù)列{an}稱為“斐波那契數(shù)列”,則(a1a3-)(a2a4-)(a3a5-)··(a2015a2017-)=( )A.1B.-1C.2017D.-201717.已知數(shù)列{a}中,a=-1,231(∈N),求數(shù)列{a}的通項(xiàng)公式.n1n+1n*n答案：1.B由已知得,數(shù)列可寫成,,故通項(xiàng)為-.2.A由Sn=2(an-1),得a1=2(a1-1),即a1=2,又a1+a2=2(a2-1),因此a2=4.3.D由n+1n,得n+2n+11,兩式相減得an+2n1,令2,得421a+a=na+a=n+-a=n=a-a=.4.B由于10,n+1n21,因此2011,3134,4459,故數(shù)列{an}的一個(gè)通項(xiàng)公式為a=a=a+n-a=+=a=+=a=+=n(1)2a=n-.5.A∵a1=,an=1-(n≥2,且n∈N*),∴a2=1-=1-=-1,-a3=1-=1--=2,411,依此類推,可得an+3n,∴a2018672×3+221,應(yīng)選A∴a=-=-=a=a=a=-.22,.S=naS=n+an+1nnn+1因此an+1=(n+1)2an+1-n2an,化簡得(n+2)an+1=nan,即,因此9·1×1a=··a=××=.7.an=n+1∵++-=an-2(n≥2),①n+12(≥2),②++=a-n②-①得=an+1-an,整理得,∴=1,又=1,∴數(shù)列是以1為首項(xiàng),1為公比的等比數(shù)列,即常數(shù)列n1,∴a=n+1.8.5或6由題意令-,,-∴)),)),解得,∴n=5或n=6.9.2n∵-an+1n20,a-=(an+1+an)(an+1-2an)=0.∵數(shù)列{an}的各項(xiàng)均為正數(shù),an+1+an>0,an+1-2an=0,即an+1=2an(n∈N*),數(shù)列{n是以2為公比的等比數(shù)列12,nn∴a.∵a=∴a=.10.解(1)a=S=a(a+1),a>0,解得a=1.111111*n∈N,an+1=Sn+1-Sn=an+1(an+1+1)-an(an+1),移項(xiàng)整理并因式分解得(an+1-an-1)(an+1+an)=0,由于{an}是正項(xiàng)數(shù)列,因此an+1+an>0,因此an+1-an-1=0,an+1-an=1.因此{(lán)a}是首項(xiàng)a=1、公差為1的等差數(shù)列,因此a=n.n1n(2)由(1)得S=a(a+1)=n(n+1),b=,T=b+b++b=--)+--.11.C∵an+1=an-an-1(n≥2),a1=m,a2=n,a3=n-m,a4=-m,a5=-n,a6=m-n,a7=m,a8=n,,an+6=an.則S2017=S336×6+1=336×(a1+a2++a6)+a1=336×0+m=m.*12.D由題意知f(Sn+2)=f(an)+f(3)=f(3an)(n∈N),兩式相減,得2an=3an-1(n≥2),則(n≥2).-又n=1時(shí),S1+2=3a1=a1+2,a1=1.-∴數(shù)列{an}是首項(xiàng)為

1,公比為的等比數(shù)列

.∴an=

.13.66

由題得

,這個(gè)數(shù)列各項(xiàng)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,∴a-a=-1,a-a1,=-1,,a-a1,10個(gè)式子之和為0,5192143620321,542,763,,19189,9個(gè)式子之和為)45a-a=a-a=a-a=a-a==.累加得a45.又1,故46,故答案為4620112015.2n-1當(dāng)n≥2時(shí),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{1}是以首項(xiàng)為112,公比為2的等比數(shù)列,nan+1=2·2n-1=2n,nan=2-1.16.B1312121,2413221,352×5-321,,∵aa-=×-=aa-=×-=-aa-==aa-=.201520171∴(13)(24)(35)··(20152017-)11008×(-1)10071aa-aa-aa-aa==-.17.解∵an+12n31(∈N*),①=a+n-n11,a=-a2=0.當(dāng)n≥2時(shí),an=2an-1+3n-4,②由①-②可得an+1-an=2an-2an-1+3,即an+1-a