2022年高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練28 數(shù)列的概念與表示 理 北師大版

2022年高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練28 數(shù)列的概念與表示 理 北師大版1.下列數(shù)列中,既是遞增數(shù)列又是無(wú)窮數(shù)列的是()A.1,B.-1,-2,-3,-4,C.- 1,-,-,-,D.1,2.數(shù)列1,的一個(gè)通項(xiàng)公式an=()A.B.C.D.3.已知數(shù)列an的前n項(xiàng)和為Sn,若Sn=2an-4(nN+),則an=()A.2n+1B.2nC.2n-1D.4.已知數(shù)列an滿(mǎn)足a1+a2+an=2a2(n=1,2,3,),則()A.a1<0B.a1>0C.a1a2D.a2=05.已知數(shù)列an的前n項(xiàng)和為Sn,a1=2,Sn=an(nN+),則S10為()A.50B.55C.100D.1106.已知數(shù)列an的首項(xiàng)a1=1,其前n項(xiàng)和Sn=n2an(nN+),則a9=()A.B.C.D.7.在數(shù)列an中,a1=1,Sn=an,則an=. 8.數(shù)列an的前n項(xiàng)和為Sn.若S2=4,an+1=2Sn+1,nN+,則S5=. 9.在數(shù)列an中,a1=0,an+1=,則S2 019=. 10.數(shù)列an的通項(xiàng)公式是an=n2+kn+4.(1)若k=-5,則數(shù)列中有多少項(xiàng)是負(fù)數(shù)?n為何值時(shí),an有最小值?并求出最小值.(2)對(duì)于nN+,都有an+1>an.求實(shí)數(shù)k的取值范圍.綜合提升組11.在數(shù)列an中,若a1=2,且對(duì)任意正整數(shù)m,k,總有am+k=am+ak,則an的前n項(xiàng)和為Sn=()A.n(3n-1)B.C.n(n+1)D.12.給定數(shù)列1,2+3+4,5+6+7+8+9,10+11+12+13+14+15+16,則這個(gè)數(shù)列的一個(gè)通項(xiàng)公式是()A.an=2n2+3n-1B.an=n2+5n-5C.an=2n3-3n2+3n-1D.an=2n3-n2+n-213.已知數(shù)列an的前n項(xiàng)和為Sn,若3Sn=2an-3n,則a2 018=()A.22 018-1B.32 018-6C. 2 018-D. 2 018-14.在一個(gè)數(shù)列中,如果每一項(xiàng)與它的后一項(xiàng)的和為同一個(gè)常數(shù),那么這個(gè)數(shù)列叫做等和數(shù)列,這個(gè)常數(shù)叫做該數(shù)列的公和,已知數(shù)列an是等和數(shù)列,且a1=2,公和為5,那么a18=. 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.意大利著名數(shù)學(xué)家斐波那契在研究兔子繁殖問(wèn)題時(shí),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(diǎn)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)所組成的數(shù)列an稱(chēng)為“斐波那契數(shù)列”,則(a1a3-)(a2a4-)(a3a5-)(a2 015a2 017-)=()A.1B.-1C.2 017D.-2 01717.(2018衡水中學(xué)二調(diào),10)數(shù)列an滿(mǎn)足a1=,an+1-1=an(an-1)(nN+),且Sn=+,則Sn的整數(shù)部分的所有可能值構(gòu)成的集合是()A.0,1,2B.0,1,2,3C.1,2D.0,2參考答案課時(shí)規(guī)范練28數(shù)列的概念與表示1.CA項(xiàng)中,數(shù)列1, , , ,是遞減數(shù)列,不符合題意;B項(xiàng)中,數(shù)列-1,-2,-3,-4,是遞減數(shù)列,不符合題意;C項(xiàng)中,數(shù)列-1,-,-, -,是遞增數(shù)列又是無(wú)窮數(shù)列,符合題意;D項(xiàng)中,數(shù)列1,是有窮數(shù)列,不符合題意,故選C.2.B由已知得,數(shù)列可寫(xiě)成, , ,故通項(xiàng)為.3.A當(dāng)n2時(shí),由Sn=2an-4,得Sn-1=2an-1-4,兩式相減得an=2an-2an-1,an=2an-1.因此數(shù)列an為公比為2的等比數(shù)列,又a1=S1=2a1-4,則a1=4,所以an=4×2n-1=2n+1.4.D根據(jù)條件Sn=a1+a2+a3+an=2a2,Sn-1=a1+a2+a3+an-1=2a2,故兩式做差得an=0,故數(shù)列的每一項(xiàng)都為0,故選D.5.D依題意Sn=(Sn-Sn-1),化簡(jiǎn)得=,故S10=····S1=××××××2=110.6.B由Sn=n2an,得Sn+1=(n+1)2an+1,所以an+1=(n+1)2an+1-n2an,化簡(jiǎn)得(n+2)an+1=nan,即=,所以a9=····a1=××××××1=.7.由題設(shè)知,a1=1.當(dāng)n2時(shí),an=Sn-Sn-1=an-an-1.=,=,=,=,=3.以上(n-1)個(gè)式子的等號(hào)兩端分別相乘,得=.a1=1,an=.8.121由于解得a1=1.由an+1=Sn+1-Sn=2Sn+1,得Sn+1=3Sn+1,所以Sn+1+=3Sn+,所以是以為首項(xiàng),3為公比的等比數(shù)列,所以Sn+=×3n-1,即Sn=,所以S5=121.9.0a1=0,an+1=,a2=,a3=-,a4=0,即數(shù)列an的取值具有周期性,周期為3,且a1+a2+a3=0,則S2 019=S3×673=0.10.解 (1)由n2-5n+4<0,解得1<n<4.nN+,n=2,3,數(shù)列中有兩項(xiàng)a2,a3是負(fù)數(shù).an=n2-5n+4=n-2-,當(dāng)n=2或n=3時(shí),an有最小值,a2=a3=-2.(2)由an+1>an知該數(shù)列是一個(gè)遞增數(shù)列,又an=n2+kn+4,可以看作是關(guān)于n的二次函數(shù),考慮到nN+,所以-<,即得k>-3.11.C遞推關(guān)系am+k=am+ak中,令k=1,得am+1=am+a1=am+2,即am+1-am=2恒成立,據(jù)此可知,該數(shù)列是一個(gè)首項(xiàng)a1=2,公差d=2的等差數(shù)列,其前n項(xiàng)和為Sn=na1+d=2n+×2=n(n+1).12.C當(dāng)n=1時(shí),a1=1,代入四個(gè)選項(xiàng),排除A、D;當(dāng)n=2時(shí),a2=9,代入B、C選項(xiàng),B、C都正確;當(dāng)n=3時(shí),a3=35,代入B、C選項(xiàng),B錯(cuò)誤,C正確,所以選C.13.A由題意可得3Sn=2an-3n,3Sn+1=2an+1-3 (n+1),兩式作差可得3an+1=2an+1-2an-3,即an+1=-2an-3,則an+1+1=-2(an+1),結(jié)合3S1=2a1-3=3a1可得a1=-3,a1+1=-2,則數(shù)列an+1是首項(xiàng)為-2,公比為-2的等比數(shù)列,據(jù)此有a2 018+1=(-2)×(-2)2 017=22 018,a2 018=22 018-1.故選A.14.3由題意得an+an+1=5an+2+an+1=5an=an+2,所以a18=a2=5-a1=3.15.2n-1當(dāng)n2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an+1是以首項(xiàng)為a1+1=2,公比為2的等比數(shù)列,an+1=2·2n-1=2n,an=2n-1.16.Ba1a3-=1×2-12=1,a2a4-=1×3-22=-1,a3a5-=2×5-32=1,a2 015a2 017-=1.(a1a3-)(a2a4-)(a3a5-)··(a2 015a2 017-)=11 008×(-1)1 007=-1.17.A對(duì)an+1-1=an(an-1)兩邊取倒數(shù),得-=,Sn=+=-+-+-=3-,由an+1-an=0,an+1an,an為遞增數(shù)列,a1=,a2=,a3=,其中S1=,整數(shù)部分為0,S2=3-=,整數(shù)部分為0,S3=,整數(shù)部分為1,由于Sn<3,故選A.