課時(shí)作業(yè)-第1節(jié)-數(shù)列的概念

第1節(jié)數(shù)列的概念知識(shí)點(diǎn)、方法基礎(chǔ)鞏固練綜合運(yùn)用練應(yīng)用創(chuàng)新練由數(shù)列的前幾項(xiàng)歸納通項(xiàng)公式1,7an與Sn的關(guān)系4,8,9數(shù)列的遞推關(guān)系2,311數(shù)列的性質(zhì)5,613綜合問(wèn)題10,12,14151.若數(shù)列的前4項(xiàng)分別是12,-13,14A.(-1)nC.(-1)nn2.若數(shù)列{an}滿足a1=1,an+1-an-1=2n,則an等于(A)A.2n+n-2 B.2n-1+n-1C.2n+1+n-4 D.2n+1+2n-2解析:因?yàn)閍n+1-an=2n+1,所以a2-a1=21+1,a3-a2=22+1,a4-a3=23+1,…,an-an-1=2n-1+1(n≥2),以上各式相加得an-a1=21+…+2n-1+(n-1)=2(1-所以an=2n+n-2.故選A.3.已知數(shù)列{an}滿足a1=1,an+1=an2-2an+1(n∈N*),則aA.1 B.0 C.2022 D.-2022解析:因?yàn)閍1=1,所以a2=(a1-1)2=0,a3=(a2-1)2=1,a4=(a3-1)2=0,…,可知數(shù)列{an}是以2為周期的數(shù)列,所以a2022=a2=0.故選B.4.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=2,Sn+1=2Sn-1(n∈N*),則a10等于(B)A.128 B.256 C.512 D.1024解析:因?yàn)镾n+1=2Sn-1(n∈N*),n≥2時(shí),Sn=2Sn-1-1,所以an+1=2an.n=1時(shí),a1+a2=2a1-1,a1=2,a2=1.所以數(shù)列{an}從第二項(xiàng)開(kāi)始為等比數(shù)列,公比為2.則a10=a2×28=1×28=256.故選B.5.(多選題)在數(shù)列{an}中,an=(n+1)(78)n,則數(shù)列{aA.第6項(xiàng) B.第7項(xiàng)C.第8項(xiàng) D.第9項(xiàng)解析:假設(shè)an最大,則有a即(所以n即6≤n≤7,所以最大項(xiàng)為第6項(xiàng)和第7項(xiàng).故選AB.6.設(shè)數(shù)列{an}的通項(xiàng)公式為an=n2-bn,若數(shù)列{an}是單調(diào)遞增數(shù)列,則實(shí)數(shù)b的取值范圍是(C)A.(-∞,-1] B.(-∞,2]C.(-∞,3) D.(-∞,92解析:因?yàn)閿?shù)列{an}是單調(diào)遞增數(shù)列,所以an+1-an=2n+1-b>0(n∈N*),所以b<2n+1(n∈N*),所以b<(2n+1)min=3,即b<3.故選C.7.已知數(shù)列32,54,76,9m-解析:由數(shù)列的前3項(xiàng)的規(guī)律可知m解得m故實(shí)數(shù)對(duì)(m,n)為(192,3答案:(192,8.若數(shù)列{an}的前n項(xiàng)和Sn=3n2-2n+1,則數(shù)列{an}的通項(xiàng)公式an=.

解析:當(dāng)n=1時(shí),a1=S1=3×12-2×1+1=2.當(dāng)n≥2時(shí),an=Sn-Sn-1=3n2-2n+1-[3(n-1)2-2(n-1)+1]=6n-5,顯然當(dāng)n=1時(shí),不滿足上式.故數(shù)列{an}的通項(xiàng)公式為an=2答案:29.已知數(shù)列{an}中,a1=1,前n項(xiàng)和Sn=n+23a(1)求a2,a3;(2)求{an}的通項(xiàng)公式.解:(1)由S2=43a2得3(a1+a2)=4a2解得a2=3a1=3.由S3=53a3得3(a1+a2+a3)=5a3解得a3=32(a1+a2(2)由題設(shè)知,當(dāng)n≥2時(shí),有an=Sn-Sn-1=n+23an-n+1整理得anan因此anan-1·an-1an-2·…·a3化簡(jiǎn)得an=(n+1)n2×1·當(dāng)n=1時(shí),a1=1滿足上式,所以{an}的通項(xiàng)公式為an=n(n+110.設(shè)數(shù)列{an}中a1=a2=1,且滿足a2n+1=3a2n-1與a2n+2-a2n+1=a2n,則數(shù)列{an}的前12項(xiàng)的和為(C)A.364 B.728C.907 D.1635解析:數(shù)列{an}中a1=a2=1,且滿足a2n+1=3a2n-1,則a3=3a1=3,a5=3a3=9,a7=3a5=27,a9=3a7=81,a11=3a9=243.由于a2n+2-a2n+1=a2n,所以a2n+2=a2n+1+a2n,故a4=a3+a2=4,a6=a5+a4=13,a8=a7+a6=40,a10=a9+a8=121,a12=a11+a10=364,所以數(shù)列{an}的前12項(xiàng)的和為1+1+3+4+9+13+27+40+81+121+243+364=907.故選C.11.已知數(shù)列{an}滿足an+1-anA.45 B.45-1C.8 D.9解析:由an+1-an=2n知a2-a1=2×1,a3-a2=2×2,…,an-an-1=2(n-1),n≥2,以上各式相加得an-a1=n2-n,n≥2,所以an=n2-n+20,n≥2,當(dāng)n=1時(shí),a1=20符合上式,所以ann=n+20n所以n≤4時(shí)ann單調(diào)遞減,n≥5時(shí)因?yàn)閍44=所以ann的最小值為a412.某數(shù)學(xué)大會(huì)會(huì)徽的主體圖案是由一連串直角三角形演化而成的(如圖),其中OA1=A1A2=A2A3=…=A7A8=1,記OA1,OA2,OA3,…,OA8的長(zhǎng)度構(gòu)成的數(shù)列為{an}(n∈N*,n≤8),則{an}的通項(xiàng)公式an=.(n∈N*,n≤8)

解析:根據(jù)題意OA1=A1A2=A2A3=…=A7A8=1,所以an2=an所以{an所以an2=n,an=答案:n13.已知數(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)若對(duì)于n∈N*,都有an+1>an,求實(shí)數(shù)k的取值范圍.解:(1)由n2-5n+4<0,解得1<n<4.因?yàn)閚∈N*,所以n=2,3,所以數(shù)列中有兩項(xiàng)是負(fù)數(shù),即為a2,a3.因?yàn)閍n=n2-5n+4=(n-52)2-9由二次函數(shù)的性質(zhì),得當(dāng)n=2或n=3時(shí),an有最小值,其最小值為a2=a3=-2.(2)由an+1>an可得(n+1)2+k(n+1)+4>n2+kn+4,整理得k>-2n-1,且對(duì)任意的n∈N*恒成立,所以k∈(-3,+∞),所以實(shí)數(shù)k的取值范圍為(-3,+∞).14.已知數(shù)列{an}中,a1=1,其前n項(xiàng)和為Sn,且滿足2Sn=(n+1)an(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)記bn=3n-λan2,若數(shù)列{b解:(1)因?yàn)?Sn=(n+1)an,所以2Sn+1=(n+2)an+1,所以2an+1=(n+2)an+1-(n+1)an,即nan+1=(n+1)an,所以an+1n所以ann=an所以an=n(n∈N*).(2)由(1)得,bn=3n-λn2.bn+1-bn=3n+1-λ(n+1)2-(3n-λn2)=2·3n-λ(2n+1).因?yàn)閿?shù)列{bn}為遞增數(shù)列,所以2·3n-λ(2n+1)>0,即λ<2·令cn=2·則cn+1cn=2·所以數(shù)列{cn}為遞增數(shù)列,所以λ<c1=2,即實(shí)數(shù)λ的取值范圍為(-∞,2).15.大衍數(shù)列