人教版導(dǎo)與練總復(fù)習(xí)數(shù)學(xué)一輪課時(shí)作業(yè)：第五章第1節(jié)　數(shù)列的概念

1、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4,8,916數(shù)列的遞推關(guān)系2,312數(shù)列的性質(zhì)5,614綜合問(wèn)題10,11,13,1517,181.若數(shù)列的前4項(xiàng)分別是12,-13,14,-15,則此數(shù)列的一個(gè)通項(xiàng)公式為(A)A.(-1)n+1n+1 B.(-1)nn+1C.(-1)nn D.(-1)n-1n2.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=2

2、2+1,a4-a3=23+1,an-an-1=2n-1+1(n2),以上各式相加得an-a1=21+2n-1+(n-1)=2(1-2n-1)1-2+n-1=2n+n-3.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(nN*),則a2 022等于(B)A.1 B.0C.2 022D.-2 022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a2 022=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(nN*),則a10等于(

3、B)A.128B.256C.512D.1 024解析:因?yàn)镾n+1=2Sn-1(nN*),n2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=a228=128=256.故選B.5.(多選題)在數(shù)列an中,an=(n+1)(78)n,則數(shù)列an中的最大項(xiàng)可以是(AB)A.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anan+1,anan-1,即(n+1)(78)n(n+2)(78)n+1,(n+1)(78)nn(78)n-1,所以n+178(n+2),78(n+1)n,即

4、6n7,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.(-,-1B.(-,2C.(-,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-b0(nN*),所以b2n+1(nN*),所以b(2n+1)min=3,即b3.故選C.7.已知數(shù)列32,54,76,9m-n,m+n10,根據(jù)前3項(xiàng)給出的規(guī)律,實(shí)數(shù)對(duì)(m,n)為.解析:由數(shù)列的前3項(xiàng)的規(guī)律可知m-n=8,m+n=11,解得m=192,n=32,故實(shí)數(shù)對(duì)(m,n)為(192,32).答案:(192,32)8.若數(shù)列an的

5、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=312-21+1=2.當(dāng)n2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,n=1,6n-5,n2.答案:2,n=1,6n-5,n29.已知數(shù)列an中,a1=1,前n項(xiàng)和Sn=n+23an.(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)=6.(2)由題設(shè)知,

6、當(dāng)n2時(shí),有an=Sn-Sn-1=n+23an-n+13an-1,整理得anan-1=n+1n-1,因此anan-1an-1an-2a3a2a2a1=n+1n-1nn-24231,化簡(jiǎn)得an=(n+1)n21a1=n(n+1)2,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+1)2(nN*).10.數(shù)列xn滿足xn+1=|xn-xn-1|(n2,nN),x1=1,x2=a(aR,a0),xn+T=xn,當(dāng)T取最小值時(shí),該數(shù)列的前2 021項(xiàng)和是(D)A.673 B.674C.1 347D.1 348解析:若T=1,則xn為常數(shù)列,故x2=x1=1,此時(shí)x3=0 x1,故T=1

7、舍去.若T=2,則x3=x1=1,故|a-1|=1,故a=2或a=0(舍去).故x4=|1-2|=1,但x5=|1-1|=0 x3,故T=2舍去.若T=3,則x3=|a-1|,x4=|a-|a-1|=x1=1,x5=|1-|1-a|=x2=a,若a1,則|a-(a-1)|=1,且|1-(a-1)|=a,整理得到|2-a|=a,解得a=1.若0aan,求實(shí)數(shù)k的取值范圍.解:(1)由n2-5n+40,解得1nan可得(n+1)2+k(n+1)+4n2+kn+4,整理得k-2n-1,且對(duì)任意的nN*恒成立,所以k(-3,+),所以實(shí)數(shù)k的取值范圍為(-3,+).15.已知數(shù)列an中,a1=1,其前

8、n項(xiàng)和為Sn,且滿足2Sn=(n+1)an(nN*).(1)求數(shù)列an的通項(xiàng)公式;(2)記bn=3n-an2,若數(shù)列bn為遞增數(shù)列,求實(shí)數(shù)的取值范圍.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+1=ann,所以ann=an-1n-1=a11=1,所以an=n(nN*).(2)由(1)得,bn=3n-n2.bn+1-bn=3n+1-(n+1)2-(3n-n2)=23n-(2n+1).因?yàn)閿?shù)列bn為遞增數(shù)列,所以23n-(2n+1)0,即1.所以數(shù)列cn為遞增數(shù)列,所以n

9、xn+1-n(n+1)2解析:由xn+1xn+xn+22得xn+1-xnxn+2-xn+1,設(shè)xn+1-xn=dn,則d1d2dn,x10-x1=d1+d2+d99d1=9(x2-x1),故A錯(cuò)誤.取x1=10,x10=1,故B錯(cuò)誤.當(dāng)k=3時(shí),數(shù)列10,9,8,7,6,5不滿足,故C錯(cuò)誤.對(duì)于D,由xn+1-xn=nn+10,知xn遞增,x1xn+1,-Sn=x1+2(x2-x1)+3(x3-x2)+n(xn-xn-1)+(n+1)(xn+1-xn)-(n+1)xn+1=x1+1+2+n-(n+1)xn+1nxn+1-n(n+1)2,故D正確.故選D.17.已知數(shù)列an滿足:(an+1+1)

10、2an+1=(an+2)2an(nN*),則下列選項(xiàng)正確的是(D)A.當(dāng)0ananB.當(dāng)an1時(shí),an+13n+18D.當(dāng)a1=4時(shí),an+1+1an+12n+2解析:對(duì)于A,由于0an1,0an+1(an+1)2an,又由函數(shù)f(x)=(x+1)2x=x2+2x+1x=x+1x+2,當(dāng)x(0,1)時(shí),函數(shù)單調(diào)遞減,可得f(an+1)f(an),所以an+11,an+11,且f(an+1)f(an),由f(x)=(x+1)2x=x2+2x+1x=x+1x+2在(1,+)上單調(diào)遞增,可得an+1an,所以B錯(cuò)誤.對(duì)于C,D,由于(an+1+1)2an+1=(an+2)2an,可得an+1+1an+1=an+1an+2+3an,當(dāng)a1=14,n=1時(shí),可得a2+1a2=a1+4a1+2=14+180時(shí),可得an0,從而an+1+1an+1an+1an+2,利用疊加法,可得an+1+1an+1a1+1a1+2n,故當(dāng)a1=4時(shí),an+1+1an+12n+2,所以D正確.故選D.18.大衍數(shù)列來(lái)源于中國(guó)古代著作乾坤譜中對(duì)易系辭上“大衍之?dāng)?shù)五十”的推論,其前10項(xiàng)為0,2,4,8,12,18,24,32,4