2019-2020年高考數(shù)學(xué)一輪復(fù)習(xí)第五章數(shù)列課時(shí)分層作業(yè)三十5.1數(shù)列的概念與簡(jiǎn)單表示法理.doc

2019-2020年高考數(shù)學(xué)一輪復(fù)習(xí)第五章數(shù)列課時(shí)分層作業(yè)三十5.1數(shù)列的概念與簡(jiǎn)單表示法理一、選擇題(每小題5分,共35分)1.已知數(shù)列的前4項(xiàng)為2,0,2,0,則歸納該數(shù)列的通項(xiàng)不可能是()A.an=(-1)n-1+1B.an=C.an=2sinD.an=cos(n-1)+1【解析】選C.對(duì)于C,當(dāng)n=3時(shí),sin =-1,則a3=-2,與題意不符.2.已知數(shù)列,2,則2是這個(gè)數(shù)列的()A.第6項(xiàng)B.第7項(xiàng)C.第19項(xiàng)D.第11項(xiàng)【解析】選B.數(shù)列即:,據(jù)此可得數(shù)列的通項(xiàng)公式為:an=,由=2,解得:n=7,即2是這個(gè)數(shù)列的第7項(xiàng).3.設(shè)數(shù)列an滿足a1=1,a2=3,且2nan=(n-1)an-1+(n+1)an+1,則a20的值是()A.4B.4C.4D.4【解析】選D.由題知:an+1=,a3=,a4=4,a5=,a6=,故an=,所以a20=4.【變式備選】數(shù)列an滿足:a1=1,且當(dāng)n2時(shí),an=an-1,則a5=()A. B.C.5D.6【解析】選A.因?yàn)閍1=1,且當(dāng)n2時(shí),an=an-1,則=,所以a5=a1=1=.4.(xx西安模擬)設(shè)數(shù)列an的前n項(xiàng)和Sn=n2,則a8的值為()A.15B.16C.49D.64【解析】選A.a8=S8-S7=82-72=15.5.(xx銀川模擬)已知數(shù)列an的通項(xiàng)公式是an=n2+kn+2,若對(duì)所有的nN*,都有an+1>an成立,則實(shí)數(shù)k的取值范圍是()A.(0,+)B.(-1,+)C.(-2,+)D.(-3,+)【解析】選D.an+1>an,即(n+1)2+k(n+1)+2>n2+kn+2,則k>-(2n+1)對(duì)所有的nN*都成立,而當(dāng)n=1時(shí),-(2n+1)取得最大值-3,所以k>-3.6.(xx北京模擬)數(shù)列an滿足an+1(an-1-an)=an-1(an-an+1),若a1=2,a2=1,則a20= ()A.B.C.D.【解題指南】數(shù)列an滿足an+1(an-1-an)=an-1(an-an+1),展開化為+=,利用等差數(shù)列的通項(xiàng)公式得出.【解析】選C.數(shù)列an滿足an+1(an-1-an)=an-1(an-an+1),展開化為+=.所以數(shù)列是等差數(shù)列,公差為-=,首項(xiàng)為.所以=+19=10,解得a20=.7.(xx黃岡模擬)已知數(shù)列an的前n項(xiàng)和Sn=n2-2n+2,則數(shù)列an的通項(xiàng)公式為()A.an=2n-3B.an=2n+3C.an=D.an=【解析】選C.當(dāng)n=1時(shí),a1=S1=1;當(dāng)n2時(shí),an=Sn-Sn-1=2n-3,由于n=1時(shí)a1的值不適合n2的解析式,故通項(xiàng)公式為an=【巧思妙解】選C.當(dāng)n=1時(shí),a1=S1=1,A、B選項(xiàng)不合題意.又a2=S2-a1=1,所以D選項(xiàng)不合題意.二、填空題(每小題5分,共15分)8.(xx長(zhǎng)沙模擬)在數(shù)列an中,a1=1,an+1=(-1)n(an+1),記Sn為an的前n項(xiàng)和,則S2 018=_ _.【解析】因?yàn)閿?shù)列an滿足a1=1,an+1=(-1)n(an+1),所以a2=-(1+1)=-2,a3=-2+1=-1,a4=-(-1+1)=0,a5=0+1=1,a6=-(1+1)=-2,a7=-2+1=-1,所以an是以4為周期的周期數(shù)列,因?yàn)? 018=5044+2,所以S2 018=504(1-2-1+0)+1-2=-1 009.答案:-1 0099.傳說(shuō)古希臘畢達(dá)哥拉斯學(xué)派的數(shù)學(xué)家經(jīng)常在沙灘上面畫點(diǎn)或用小石子表示數(shù).他們研究過(guò)如圖所示的三角形數(shù):將三角形數(shù)1,3,6,10,記為數(shù)列an,則數(shù)列an的通項(xiàng)公式為_.【解析】由題干圖可知,an+1-an=n+1,a1=1,由累加法可得an=.答案:an=10.(xx合肥模擬)已知數(shù)列an的前n項(xiàng)和Sn=3-32n,nN*,則an=_.【解析】(1)當(dāng)n=1時(shí),a1=S1=3-321=-3.(2)當(dāng)n2時(shí),an=Sn-Sn-1=(3-32n)-(3-32n-1)=-32n-1.當(dāng)n=1時(shí)也符合上式,綜合(1)(2),得an=-32n-1.答案:-32n-11.(5分)(xx衡水模擬)數(shù)列an滿足a1=2,an+1=(an>0),則an=()A.B.2n-1C.4n-1D.2n【解析】選A.因?yàn)閿?shù)列an滿足a1=2,an+1=(an>0),所以log2an+1=2log2an=2,所以log2an是公比為2的等比數(shù)列,所以log2an=log2a12n-1an=.2.(5分)若數(shù)列an是正項(xiàng)數(shù)列,且+=n2+n,則a1+等于()A.2n2+2nB.n2+2nC.2n2+nD.2(n2+2n)【解析】選A.因?yàn)?=n2+n,所以n=1時(shí),=2,解得a1=4,n2時(shí),+=(n-1)2+n-1,相減可得=2n,所以an=4n2,n=1時(shí)也成立,所以=4n.則a1+=4(1+2+n)=4=2n2+2n.3.(5分)(xx鄭州模擬)意大利數(shù)學(xué)家列昂納多斐波那契以兔子繁殖為例,引入“兔子數(shù)列”:1,1,2,3,5,8,13,21,34,55,89,144,233,即F(1)=F(2)=1,F(n)=F(n-1)+ F(n-2)(n3,nN*),此數(shù)列在現(xiàn)代物理、準(zhǔn)晶體結(jié)構(gòu)、化學(xué)等領(lǐng)域都有著廣泛的應(yīng)用,若此數(shù)列被3整除后的余數(shù)構(gòu)成一個(gè)新數(shù)列bn,則b2 018=_.【解析】由題意得,引入“兔子數(shù)列”:1,1,2,3,5,8,13,21,34,55,89,144,233,.此數(shù)列被3整除后的余數(shù)構(gòu)成一個(gè)新數(shù)列為1,1,2,0,2,2,1,0,1,1,2,0,2, 2,1,0,構(gòu)成以8為周期的周期數(shù)列,所以b2 018=b2=1.答案:14.(12分)(xx泰安模擬)設(shè)數(shù)列an的前n項(xiàng)和為Sn,數(shù)列Sn的前n項(xiàng)和為Tn,Tn滿足Tn=2Sn-n2,nN*. (1)求a1的值.(2)求數(shù)列an的通項(xiàng)公式.【解析】(1)當(dāng)n=1時(shí),T1=2S1-1, 因?yàn)門1=S1=a1,所以a1=2a1-1,所以a1=1.(2)當(dāng)n2時(shí),Tn-1=2Sn-1-(n-1)2,則Sn=Tn-Tn-1=2Sn-n2-2Sn-1-(n-1)2=2(Sn-Sn-1)-2n+1=2an-2n+1,因?yàn)楫?dāng)n=1時(shí),a1=S1=1也滿足上式,所以Sn=2an-2n+1(n1),當(dāng)n2時(shí),Sn-1=2an-1-2(n-1)+1,兩式相減得an=2an-2an-1-2,所以an=2an-1+2(n2),所以an+2=2(an-1+2),因?yàn)閍1+2=30,所以數(shù)列an+2是以3為首項(xiàng),2為公比的等比數(shù)列.所以an+2=32n-1,所以an=32n-1-2,當(dāng)n=1時(shí)也成立,所以an=32n-1-2(nN*).5.(13分)(xx銀川模擬)已知函數(shù)f(x)=2x-2-x,數(shù)列an滿足f(log2an)=-2n.(1)求數(shù)列an的通項(xiàng)公式.(2)證明:數(shù)列an是遞減數(shù)列.【解析】(1)因?yàn)閒(x)=2x-,f(log2an)=-2n,所以an-=-2n,所以+2nan-1=0,解得an=-n,因?yàn)閍n>0,所以an=-n,nN*.(2)=<1,因?yàn)閍n>0,所以an+1<an,所以數(shù)列an是遞減數(shù)列.