年高考數(shù)學(xué)（理）課時(shí)作業(yè)加練一課（4）遞推數(shù)列的通項(xiàng)的求法

加練一課(四)遞推數(shù)列的通項(xiàng)的求法一、選擇題(本大題共12小題,每小題5分,共60分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.[2017·遵義航天高級(jí)中學(xué)月考]數(shù)列an滿足a1=1,an+1=2an1,則an= (A.1 B.2n1C.n D.12.已知數(shù)列{an}對(duì)任意的p,q∈N*滿足ap+q=ap+aq且a2=6,那么a10等于 ()A.165 B.33C.30 D.213.[2017·黃山二模]已知數(shù)列an的前n項(xiàng)和為Sn,且a1=2,an+1=Sn+1(n∈N*),則S5= (A.31 B.42C.37 D.474.若數(shù)列{an}前8項(xiàng)的值各異,且an+8=an對(duì)任意的n∈N*都成立,則下列數(shù)列中可取遍{an}前8項(xiàng)值的數(shù)列為 ()A.{a2k+1} B.{a3k+1}C.{a4k+1} D.{a6k+1}5.[2017·揭陽模擬]已知數(shù)列an滿足a1=1,an+1=n+1n+2an,則an=A.1n B.C.12n-1 6.[2017·三明質(zhì)檢]已知數(shù)列an的前n項(xiàng)和為Sn,且a1=1,an+1·an=2n(n∈N*),則S2017=(A.3·210083 B.220171C.220093 D.2101037.已知數(shù)列an滿足a1=1,an+1=an+1n+1+n,則anA.3n-2 BC.n D.28.已知數(shù)列an滿足a1=1,anan+1=nanan+1(n∈N*),則an= (A.2n2+1 C.1n2-n9.[2017·贛州期末]已知數(shù)列{an}的前n項(xiàng)和為Sn,若an+1+(1)nan=n,則S40= ()A.120 B.150C.210 D.42010.已知數(shù)列{an}滿足a1=2,且an=2nan-1an-1+n-1(n≥2,n∈NA.2n+12nC.n·2n211.[2017·福州第一中學(xué)質(zhì)檢]已知數(shù)列an滿足a1=a2=12,an+1=2an+an1(n∈N*,n≥2),則∑i=22017A.0 B.1C.2 D.312.已知Sn為數(shù)列an的前n項(xiàng)和,且an=2an-1,n≥6,an-1+1,2≤n<6,a1=a(a∈R).給出下列3個(gè)結(jié)論:①數(shù)列an+5一定是等比數(shù)列;②若S5<100,則a<18;③若a3,a6,A.②B.②③C.①③D.①②③二、填空題(本大題共4小題,每小題5分,共20分.把答案填在題中橫線上)13.若數(shù)列an滿足a1=1,an+1=an+2,則a10=14.[2017·深圳調(diào)研]若數(shù)列an,bn滿足a1=b1=1,bn+1=an,an+1=3an+2bn,n∈N*,則a2017a2016=15.[2017·株洲一模]已知數(shù)列{an}滿足a1=1,an=a1+2a2+3a3+…+(n1)an1(n≥2),則數(shù)列{an}的通項(xiàng)公式為an=.

16.[2018·南寧二中、柳州高中聯(lián)考]已知數(shù)列2008,2009,1,2008,…若這個(gè)數(shù)列從第二項(xiàng)起,每一項(xiàng)都等于它的前后兩項(xiàng)之和,則這個(gè)數(shù)列的前2018項(xiàng)之和S2018=.

加練一課(四)1.A[解析]∵an+1=2an1,∴an+11=2(an1).∵a11=0,∴an1=0,即an=1,故選A.2.C[解析]a4=a2+a2=12,a6=a4+a2=18,a10=a6+a4=30.故選C.3.D[解析]由an+1=Sn+1①,可得an=Sn1+1(n≥2)②,①②得an+1=2an,又∵a2=S1+1=3,a1=2,∴S5=2+3×(1-24)1-2=474.B[解析]因?yàn)閿?shù)列{an}前8項(xiàng)的值各異,且an+8=an對(duì)任意的n∈N*都成立,所以該數(shù)列為周期為8的周期數(shù)列.為使數(shù)列中可取遍{an}前8項(xiàng)的值,必須保證項(xiàng)數(shù)被8除的余數(shù)可以取到0,1,2,3,4,5,6,7.經(jīng)驗(yàn)證A,C,D都不可以,因?yàn)樗鼈兊捻?xiàng)數(shù)全部由奇數(shù)組成,被8除的余數(shù)只能是奇數(shù),故選B.5.B[解析]由條件知an+1an=n+1n+2,分別令n=1,2,3,…,(n1)(n≥2),可得a2a1=23,a3a2=34,a4a3=45,…,anan-1=nn+1,累乘得a2a1·a3a2·a4a36.D[解析]∵數(shù)列an滿足a1=1,an+1·an=2n(n∈N*),∴a2·a1=2,解得a2=2.由題得an+1anan+2an+1=2n2n+1,即an+2an=2,∴數(shù)列{an}的奇數(shù)項(xiàng)與偶數(shù)項(xiàng)分別成等比數(shù)列,首項(xiàng)分別為a1=1,a2=2,公比都為2,則S2017=(a1+a3+…+a2017)+(a27.C[解析]由條件知an+1an=1n+1+n=n+1n.分別令n=1,2,3,…,(n1),代入上式得到(n1)個(gè)等式,這些等式累加可得(a2a1)+(a3a2)+(a4a3)+…+(anan1)=(21)+(32)+(43)+…+(nn-1),即ana1=n1.又因?yàn)閍1=1,所以an8.D[解析]因?yàn)閍nan+1=nanan+1,所以an-an+1anan+1=1an+11an=n,所以1an=1an1an-1+1an-11an-2+…+1a21a1+1a1=(n1)9.D[解析]由已知得a3+a1=(a3+a2)(a2a1)=1,同理可得a5+a7=1,…,a37+a39=1,又a2+a4=(a3+a2)+(a4a3)=2+3=5,a6+a8=13,…,a38+a40=77,∴S40=(a1+a3+…+a39)+(a2+a4+…+a40)=10×1+(5+13+…+77)=10+410=420,故選D.10.C[解析]由an=2nan-1an-1+n-1,得nan=n-12an-1+12,于是nan1=12n-1an-11(n≥2,n∈N*).又1a11=12,∴數(shù)列nan1是以12為首項(xiàng),12為公比的等比數(shù)列,故nan1=12n(n≥2,n∈N11.B[解析]∵a1=12,a2=12,an+1=2an+an1,∴an+1-an-12an=1,a3=2a2+a1=32,∴1ai-1ai+1=1ai-1ai+1·ai+1-ai-12ai=12ai1ai-11ai+1=121ai-1ai1aiai+1,∑i=220171ai12.B[解析]根據(jù)題意,數(shù)列an滿足an=2an-1,n≥6,an-1+1,2≤n<6,且a1=a,則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,…對(duì)于①,當(dāng)a=4時(shí),a6=2a+8=0,此時(shí)數(shù)列an+5不是等比數(shù)列,故①錯(cuò)誤;對(duì)于②,若S5<100,則有S5=(a1+a2+…+a5)=5(a+2)<100,則有a<18,故②正確;對(duì)于③,根據(jù)題意,a3=a+2,a6=2a+8,a9=24a5=16×(a+4),若a3,a6,a9成等比數(shù)列,則有(2a+8)2=(a+2)×16×(a+4),且a6=213.19[解析]因?yàn)閍n+1=an+2,所以an+1an=2,所以數(shù)列an是首項(xiàng)為1,公差為2的等差數(shù)列,所以a10=1+(101)×2=1914.22017[解析]由題得,a2=3a1+2b1=5,當(dāng)n≥2時(shí),an+1=3an+2bn=3an2an1,所以an+1an=2(anan1),又a2a1=4,所以數(shù)列{anan1}是首項(xiàng)為4,公比為2的等比數(shù)列,所以a2017a2016=4×220161=22017.15.1,n=1,n!2,n≥2[解析]當(dāng)n≥2時(shí),由已知得an+1=a1+2a2+3a3+…+(n1)an1+nan,用此等式減去已知等式,得an+1an=nan,即an+1=(n+1)an,又a2=a1=1,∴a1=1,a2a1=1,a3a2=3,a4a3=4,…,anan-1=n,將以上n個(gè)式子相乘,得a