课时作业(二十八)1.C解析由题意,得a2=2a1+1=3,a3=2a2+1=7,a4=2a3+1=15,故选C.2.B解析易知数列-1,4,-9,16,-25,的一个通项公式为an=-1nn2,故选B.3.B解析观察数列,得出规律:a2-a1=21,a3-a2=22,a4-a3=23,a5-a4=24,因此a6-a5=25,所以a6=62,故选B.4.5解析因为an=3n2-28n=3n-1432-1963,且nN*,所以当n=5时,an取得最小值.5.2n解析当n2时,an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n.当n=1时,a1=S1=2,满足上式.故an=2n.6.C解析当n2时,Sn=n+23an,Sn-1=n+13an-1.两式作差可得an=Sn-Sn-1=n+23an-n+13an-1,则anan-1=n+1n-1=1+2n-1,据此可得,当n=2时,anan-1取到最大值3.7.A解析(3n+2)an+1=(3n-1)an,an+1=3n-13n+2an,an=3(n-1)-13(n-1)+23(n-2)-13(n-2)+232-132+23-13+2a1=3n-43n-13n-73n-458253=63n-1,故选A.8.D解析根据题意可知a1=3,a2=6,an+2=an+1-an,那么a3=a2-a1=3,a4=a3-a2=-3,a5=a4-a3=-6,a6=a5-a4=-3,a7=a6-a5=3,a8=a7-a6=6,可知数列an的周期为6,那么a2018=a3366+2=a2=6,故选D.9.D解析an=8+2n-72n,an+1=8+2n-52n+1,an+1-an=2n-52n+1-2n-72n=2n-5-2(2n-7)2n+1=-2n+92n+1.当1n4时,an+1an,即a5a4a3a2a1;当n5时,an+1a6a7.因此数列an先递增后递减,当n=5时,a5=25932为最大项,即M=25932,又当n时,an8,a1=112,最小项为112,即m=112,m+M=112+25932=43532.故选D.10.D解析根据题意可构造数列an-2bn,则an+1-2bn+1=an+2bn-2an-2bn=(1-2)an-(1-2)2bn=(1-2)(an-2bn).因为a1=b1=1,所以a1-2b1=1-2,所以an-2bn是以1-2为首项,1-2为公比的等比数列,故an-2bn=(1-2)n,所以A,B不正确.因为an-2bn的公比为1-2,其绝对值小于1,所以|an-2bn|为递减数列,所以C不正确.anbn-2=1bn|an-2bn|,易知数列an,bn为递增数列,故1bn为递减数列,又|an-2bn|为递减数列,故anbn-2为递减数列,D正确.11.an=1,n=1,2n-2,n2解析由an+1=Sn,可得an=Sn-1(n2),-得an+1-an=Sn-Sn-1=an(n2),即an+1an=2(n2),又a2=S1=1,所以a2a1=12,则数列an从第二项起是以1为首项2为公比的等比数列,所以an=1,n=1,2n-2,n2.12.53,4解析因为an是递减数列,数列an从a4项开始用式子(t-13)x-3计算,所以只要t-130,即t52且a3a4,即t53且9-9t+18t-13,解得53t4.综上,t的取值范围是53t4.13.n3n-2解析由an-an+1=2anan+1n(n+1)可得1an+1-1an=2n(n+1)=21n-1n+1,利用累加法可得1an-1an-1+1an-1-1an-2+1a2-1a1=21n-1-1n+21n-2-1n-1+21-12,即1an-1a1=21-1n,可得1an=3-2n=3n-2n,即an=n3n-2.14.解:(1)设等差数列an的公差为d,由a3=24,S11=0,可得a1+2d=24,11a1+10112d=0,解得a1=40,d=-8,所以数列an的通项公式为an=48-8n.(2)由(1)知Sn=-4n2+44n=-4n-1122+121,因为nN*,所以当n=5或6时,Sn取得最大值.15.解:(1)因为a2=72,所以由a2=3a1-1可求得a1=32.因为an+1=3an-1,所以an+1-12=3an-12,所以数列an-12是以1为首项,以3为公比的等比数列.所以an-12=3n-1,即an=12+3n-1.故Sn=n2+3n-12=3n+n-12.(2)依题意,3n-1+13n-1m,即13+43(3n-1)m对任意的nN*恒成立.设cn=13+43(3n-1),则易知数列cn是递减数列,所以cnmax=c1=1.综上,可得m1.故所求实数m的取值范围是1,+).16.解:(1)证明:由题知bn+1+2bn+2=2bn+2+2bn+2=2,b1=a2-a1=4-2=2,b1+2=4,数列bn+2是以4为首项以2为公比的等比数列.(2)由(1)可得bn+2=42n-1,故bn=2n+1-2.an+1-an=bn,a2-a1=b1,a3-a2=b2,a4-a3=b3,an-an-1=bn-1,累加得an-a1=b1+b2+b3+bn-1=(22-2)+(23-2)+(24-2)+(2n-2)=22(1-2n-1)1-2-2(n-1)=2n+1-2n-2,则an=2n+1-2n.加练一课(四)1.A解析an+1=2an-1,an+1-1=2(an-1).a1-1=0,an-1=0,即an=1,故选A.2.C解析a4=a2+a2=12,a6=a4+a2=18,a10=a6+a4=30.故选C.3.D解析由an+1=Sn+1,可得an=Sn-1+1(n2),-得an+1=2an,又a2=S1+1=3,a1=2,S5=2+3(1-24)1-2=47,故选D.4.B解析因为数列an前8项的值各异,且an+8=an对任意的nN*都成立,所以该数列为周期为8的周期数列.为使数列中可取遍an前8项的值,必须保证项数被8除的余数可以取到0,1,2,3,4,5,6,7.经验证A,C,D都不可以,因为它们的项数全部由奇数组成,被8除的余数只能是奇数,故选B.5.B解析由条件知an+1an=n+1n+2,分别令n=1,2,3,(n-1)(n2),可得a2a1=23,a3a2=34,a4a3=45,anan-1=nn+1,累乘得a2a1a3a2a4a3anan-1=2334n-1nnn+1,即ana1=2n+1.又a1=1,an=2n+1,故选B.6.D解析数列an满足a1=1,an+1an=2n(nN*),a2a1=2,解得a2=2.由题得an+1anan+2an+1=2n2n+1,即an+2an=2,数列an的奇数项与偶数项分别成等比数列,首项分别为a1=1,a2=2,公比都为2,则S2017=(a1+a3+a2017)+(a2+a4+a2016)=1(1-21009)1-2+2(1-21008)1-2=21010-3,故选D.7.C解析由条件知an+1-an=1n+1+n=n+1-n.分别令n=1,2,3,(n-1),代入上式得到(n-1)个等式,这些等式累加可得(a2-a1)+(a3-a2)+(a4-a3)+(an-an-1)=(2-1)+(3-2)+(4-3)+(n-n-1),即an-a1=n-1.又因为a1=1,所以an=n,故选C.8.D解析因为an-an+1=nanan+1,所以an-an+1anan+1=1an+1-1an=n,所以1an=1an-1an-1+1an-1-1an-2+1a2-1a1+1a1=(n-1)+(n-2)+3+2+1+1a1=(n-1)(n-1+1)2+1=n2-n+22,则an=2n2-n+2.9.D解析由已知得a3+a1=(a3+a2)-(a2-a1)=1,同理可得a5+a7=1,a37+a39=1,又a2+a4=(a3+a2)+(a4-a3)=2+3=5,a6+a8=13,a38+a40=77,S40=(a1+a3+a39)+(a2+a4+a40)=101+(5+13+77)=10+410=420,故选D.10.C解析由an=2nan-1an-1+n-1,得nan=n-12an-1+12,于是nan-1=12n-1an-1-1(n2,nN*).又1a1-1=-12,数列nan-1是以-12为首项,12为公比的等比数列,故nan-1=-12n(n2,nN*),当n=1时,a1=2满足上式,则nan-1=-12n,an=n2n2n-1(nN*),故选C.11.B解析a1=12,a2=12,an+1=2an+an-1,an+1-an-12an=1,a3=2a2+a1=32,1ai-1ai+1=1ai-1ai+1ai+1-ai-12ai=12ai1ai-1-1ai+1=121ai-1ai-1aiai+1,i=220171ai-1ai+1=121a1a2-1a2a3+1a2a3-1a3a4+1a2016a2017-1a2017a2018=121a1a2-1a2017a2018=124-1a2017a2018=2-12a2017a20181,1i=220171ai-1ai+12,则i=220171ai-1ai+1的整数部分是1,故选B.12.B解析根据题意,数列an满足an=2an-1,n6,an-1+1,2n6,且a1=a,则a2=a1+1=a+1,a3=a2+1=a+2,a4=a3+1=a+3,a5=a4+1=a+4,a6=2a5=2a+8,a7=2a6,对于,当a=-4时,a6=2a+8=0,此时数列an+5不是等比数列,故错误;对于,若S5100,则有S5=(a1+a2+a5)=5(a+2)100,则有a18,故正确;对于,根据题意,a3=a+2,a6=2a+8,a9=24a5=16(a+4),若a3,a6,a9成等比数列,则有(2a+8)2=(a+2)16(a+4),且a6=2a+80,解得a=-43,故正确.故选B.13.19解析因为an+1=an+2,所以an+1-an=2,所以数列an是首项为1,公差为2的等差数列,所以a10=1+(10-1)2=19.14.22017解析由题得,a2=3a1+2b1=5,当n2时,an+1=3an+2bn=3an-2an-1,所以an+1-an=2(an-an-1),又a2-a1=4,所以数列an-an-1是首项为4,公比为2的等比数列,所以a2017-a2016=422016-1=22017.15.1,n=1,n!2,n2解析当n2时,由已知得an+1=a1+2a2+3a3+(n-1)an-1+nan,用此等式减去已知等式,得an+1-an=nan,即an+1=(n+1)an,又a2=a1=1,a1=1,a2a1=1,a3a2=3,a4a3=4,anan-1=n,将以上n个式子相乘,得an=n!2(n2).当n=1时,a