高考数学（文）课时作业加练一课（四）递推数列的通项的求法

加练一课(四)递推数列的通项的求法时间/30分钟分值/80分一、选择题(本大题共10小题,每小题5分,共50分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.在数列{an}中,已知a1=2,a2=7,an+2等于anan+1(n∈N*)的个位数,则a2018= ()A.8 B.6C.4 D.22.已知数列{an}的前n项积为n2,那么当n≥2时,an= ()A.2n1 B.n2C.(n+1)23.数列{an}定义如下:a1=1,当n≥2时,an=1+an2,n为偶数,1an-1,n为奇数.A.7 B.8C.9 D.104.[2017·河南八市三模]已知数列{an}满足an+1(an1an)=an1(anan+1),若a1=2,a2=1,则a20=()A.1210 B.C.110 D.5.已知数列{an}中,a1=1,前n项和Sn=n+23an,则数列{an}的通项公式为 (A.an=n(B.an=nC.an=n(n+1) D.an=n(n1)6.已知数列{an}的前n项和为Sn,a1=1,Sn=2an+1,则Sn= ()A.2n1 B.3C.23D.17.[2017·衡水中学模拟]已知数列{an}满足a1=2,an+1=an2(an>0),则an= (A.10n2 B.10n1C.102n-1 8.已知各项均为正数的数列{an}中,a1=1,数列的前n项和Sn满足SnSn-1Sn1Sn=2SnSn-1(n∈N*,且n≥2),则A.120 B.240C.360 D.4809.已知数列{an}满足an+2=an+1an,且a1=2,a2=3,则a2018的值为 ()A.3 B.3C.2 D.210.[2017·沈阳二中月考]设数列{an}的前n项和为Sn,且a1=1,{Sn+nan}为常数列,则an= ()A.13B.2C.6(D.5-2二、填空题(本大题共6小题,每小题5分,共30分)11.已知数列{an}满足a1=1,an=an1+2n(n≥2),则a7=.

12.已知正项数列{an}满足a1=1,a2=2,2an2=an+12+an-12(n∈N*,n≥13.[2018·合肥六中月考]已知数列{an}满足an+1=3an+1,且a1=1,则数列{an}的通项公式为an=.

14.[2017·宁波期中]已知数列{an}的前n项和Sn=n2+2n1,则数列{an}的通项公式为an=.

15.已知数列{an}满足a1=1,a2=4,an+2+2an=3an+1(n∈N*),则数列{an}的通项公式为an=.

16.[2017·兰州一诊]已知数列{an},{bn},若b1=0,an=1n(n+1),当n≥2时,有bn=bn1+an1,则b加练一课(四)递推数列的通项的求法1.D[解析]由题易得a3=4,a4=8,a5=2,a6=6,a7=2,a8=2,a9=4,a10=8,所以数列{an}中的项从第3项开始呈周期性出现,周期为6,故a2018=a335×6+8=a8=2.2.D[解析]设数列{an}的前n项积为Tn,则Tn=n2,当n≥2时,an=TnTn3.C[解析]因为a1=1,所以a2=1+a1=2,a3=1a2=12,a4=1+a2=3,a5=1a4=13,a6=1+a3=32,a7=1a6=23,a8=1+a4=4,a9=14.C[解析]将an+1(an1an)=an1(anan+1)化为1an-1+1an+1=2an,可知数列1an是等差数列,其首项为12,公差为1a21a1=12,∴1a205.A[解析]当n>1时,有an=SnSn1=n+23ann+13an1,整理得an=n+1n-1an1,则a2=31a1,a3=42a2,…,an1=nn-2an2,an=n+1n-1an1,将以上(n1)个等式等号的两端分别相乘,整理得an=n(n+1)2,6.B[解析]由Sn=2an+1,得Sn=2(Sn+1Sn),即2Sn+1=3Sn,则Sn+1Sn=32,又S1=a1=1,所以7.D[解析]因为数列{an}满足an+1=an2(an>0),所以log2an+1=2log2an⇒log2an+1log2an=2,所以{log2an}是公比为2的等比数列,则log2an=log28.D[解析]由SnSn-1Sn1Sn=2SnSn-1可得SnSn-1=2,所以{Sn}是以1为首项,2为公差的等差数列,则Sn=2n1,即Sn=(2n1)2,所以a61=S61S9.B[解析]由题意得a3=a2a1=1,a4=a3a2=2,a5=a4a3=3,a6=a5a4=1,a7=a6a5=2,a8=a7a6=3,∴数列{an}是周期为6的周期数列,∴a2018=a2=3.10.B[解析]由题意知Sn+nan=2,当n≥2时,(n+1)an=(n1)an1,从而a2a1·a3a2·a4a3·…·anan-1=13·24·…·n-1n+1,则an=11.55[解析]∵an=an1+2n(n≥2),∴anan1=2n(n≥2),则a2a1=4,a3a2=6,a4a3=8,a5a4=10,a6a5=12,a7a6=14,∴a7=1+4+6+8+10+12+14=55.12.19[解析]由2an2=an+12+an-12(n∈N*,n≥2),可得数列{an2}是等差数列,则公差d=a22a12=3,又a12=1,∴an2=1+3(13.12(3n1)[解析]因为an+1=3an+1,所以an+1+12=3an+12,所以数列an+12是以32为首项,以3为公比的等比数列,所以an+12=32×3n14.2,n=1,2n+1,n≥2[解析]∵Sn=n2+2n1,∴当n=1时,a1=1+21=2,当n≥2时,an=SnSn1=n2+2n1[(n1)2+2(n1)1]=2n+1.∵当n=1时,a1=2≠2+115.3×2n12[解析]由an+2+2an3an+1=0,得an+2an+1=2(an+1an),∴数列{an+1an}是以a2a1=3为首项,2为公比的等比数列,∴an+1an=3×2n1,∴当n≥2时,anan1=3×2n2,…,a3a2=3×2,a2a1=3,将以上各式累加得ana1=3×2n2+…+3×2+3=3(2n11),∴an=3×2n12(当n=1时,也满足此式).16.20162017[解析]