高考數學（文）課時作業(yè)加練一課（四）遞推數列的通項的求法

加練一課(四)遞推數列的通項的求法時間/30分鐘分值/80分一、選擇題(本大題共10小題,每小題5分,共50分,在每小題給出的四個選項中,只有一項是符合題目要求的)1.在數列{an}中,已知a1=2,a2=7,an+2等于anan+1(n∈N*)的個位數,則a2018= ()A.8 B.6C.4 D.22.已知數列{an}的前n項積為n2,那么當n≥2時,an= ()A.2n1 B.n2C.(n+1)23.數列{an}定義如下:a1=1,當n≥2時,an=1+an2,n為偶數,1an-1,n為奇數.A.7 B.8C.9 D.104.[2017·河南八市三模]已知數列{an}滿足an+1(an1an)=an1(anan+1),若a1=2,a2=1,則a20=()A.1210 B.C.110 D.5.已知數列{an}中,a1=1,前n項和Sn=n+23an,則數列{an}的通項公式為 (A.an=n(B.an=nC.an=n(n+1) D.an=n(n1)6.已知數列{an}的前n項和為Sn,a1=1,Sn=2an+1,則Sn= ()A.2n1 B.3C.23D.17.[2017·衡水中學模擬]已知數列{an}滿足a1=2,an+1=an2(an>0),則an= (A.10n2 B.10n1C.102n-1 8.已知各項均為正數的數列{an}中,a1=1,數列的前n項和Sn滿足SnSn-1Sn1Sn=2SnSn-1(n∈N*,且n≥2),則A.120 B.240C.360 D.4809.已知數列{an}滿足an+2=an+1an,且a1=2,a2=3,則a2018的值為 ()A.3 B.3C.2 D.210.[2017·沈陽二中月考]設數列{an}的前n項和為Sn,且a1=1,{Sn+nan}為常數列,則an= ()A.13B.2C.6(D.5-2二、填空題(本大題共6小題,每小題5分,共30分)11.已知數列{an}滿足a1=1,an=an1+2n(n≥2),則a7=.

12.已知正項數列{an}滿足a1=1,a2=2,2an2=an+12+an-12(n∈N*,n≥13.[2018·合肥六中月考]已知數列{an}滿足an+1=3an+1,且a1=1,則數列{an}的通項公式為an=.

14.[2017·寧波期中]已知數列{an}的前n項和Sn=n2+2n1,則數列{an}的通項公式為an=.

15.已知數列{an}滿足a1=1,a2=4,an+2+2an=3an+1(n∈N*),則數列{an}的通項公式為an=.

16.[2017·蘭州一診]已知數列{an},{bn},若b1=0,an=1n(n+1),當n≥2時,有bn=bn1+an1,則b加練一課(四)遞推數列的通項的求法1.D[解析]由題易得a3=4,a4=8,a5=2,a6=6,a7=2,a8=2,a9=4,a10=8,所以數列{an}中的項從第3項開始呈周期性出現,周期為6,故a2018=a335×6+8=a8=2.2.D[解析]設數列{an}的前n項積為Tn,則Tn=n2,當n≥2時,an=TnTn3.C[解析]因為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]將an+1(an1an)=an1(anan+1)化為1an-1+1an+1=2an,可知數列1an是等差數列,其首項為12,公差為1a21a1=12,∴1a205.A[解析]當n>1時,有an=SnSn1=n+23ann+13an1,整理得an=n+1n-1an1,則a2=31a1,a3=42a2,…,an1=nn-2an2,an=n+1n-1an1,將以上(n1)個等式等號的兩端分別相乘,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,則Sn+1Sn=32,又S1=a1=1,所以7.D[解析]因為數列{an}滿足an+1=an2(an>0),所以log2an+1=2log2an?log2an+1log2an=2,所以{log2an}是公比為2的等比數列,則log2an=log28.D[解析]由SnSn-1Sn1Sn=2SnSn-1可得SnSn-1=2,所以{Sn}是以1為首項,2為公差的等差數列,則Sn=2n1,即Sn=(2n1)2,所以a61=S61S9.B[解析]由題意得a3=a2a1=1,a4=a3a2=2,a5=a4a3=3,a6=a5a4=1,a7=a6a5=2,a8=a7a6=3,∴數列{an}是周期為6的周期數列,∴a2018=a2=3.10.B[解析]由題意知Sn+nan=2,當n≥2時,(n+1)an=(n1)an1,從而a2a1·a3a2·a4a3·…·anan-1=13·24·…·n-1n+1,則an=11.55[解析]∵an=an1+2n(n≥2),∴anan1=2n(n≥2),則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),可得數列{an2}是等差數列,則公差d=a22a12=3,又a12=1,∴an2=1+3(13.12(3n1)[解析]因為an+1=3an+1,所以an+1+12=3an+12,所以數列an+12是以32為首項,以3為公比的等比數列,所以an+12=32×3n14.2,n=1,2n+1,n≥2[解析]∵Sn=n2+2n1,∴當n=1時,a1=1+21=2,當n≥2時,an=SnSn1=n2+2n1[(n1)2+2(n1)1]=2n+1.∵當n=1時,a1=2≠2+115.3×2n12[解析]由an+2+2an3an+1=0,得an+2an+1=2(an+1an),∴數列{an+1an}是以a2a1=3為首項,2為公比的等比數列,∴an+1an=3×2n1,∴當n≥2時,anan1=3×2n2,…,a3a2=3×2,a2a1=3,將以上各式累加得ana1=3×2n2+…+3×2+3=3(2n11),∴an=3×2n12(當n=1時,也滿足此式).16.20162017[解析]