2020版高考數(shù)學(xué)復(fù)習(xí)第五單元專題探究3由數(shù)列的遞推關(guān)系式求通項(xiàng)公式練習(xí).docx

專題探究3由數(shù)列的遞推關(guān)系式求通項(xiàng)公式 1.2018沈陽聯(lián)考 已知數(shù)列an滿足a1=1,an=an-1+n(n2,nN*),則a5=()A.6B.10C.15D.212.已知數(shù)列an對任意的p,qN*滿足ap+q=ap+aq,且a2=-6,則a10等于()A.-165B.-33C.-30D.-213.已知數(shù)列an滿足a1=2,an+1=2an-1,則a6=()A.31B.32C.33D.344.2018鄭州模擬 已知數(shù)列an滿足an+1=anan+1,且a1=12,則a8=()A.17B.18C.19D.1105.2018河南師大附中模擬 已知數(shù)列an滿足an+1=an+12n,且a1=1,則an=.6.已知數(shù)列an滿足a1=36,an+1=an+2n,則ann的最小值為()A.10B.11C.12D.137.2018中山模擬 設(shè)函數(shù)f(x)滿足f(n+1)=3f(n)+n3(nN*),且f(1)=1,則f(18)=()A.20B.38C.52D.358.已知數(shù)列an的前n項(xiàng)和Sn=2an-1,則滿足ann2的正整數(shù)n的集合為()A.1,2B.1,2,3,4C.1,2,3D.1,2,49.2018南昌模擬 已知數(shù)列an滿足an=3an-1+3n-1(n2),且a1=5.若an+3n為等差數(shù)列,則實(shí)數(shù)=()A.2B.5C.-12D.1210.2018銅仁模擬 已知數(shù)列an滿足a1=33,an+1-an=2n,則an=.11.2018重慶江津中學(xué)、合川中學(xué)等七校聯(lián)考 在數(shù)列an中,a1=2,an+1-an=2(n+1),則數(shù)列1an的前10項(xiàng)的和為.12.2018鄭州聯(lián)考 設(shè)數(shù)列an對任意nN*都滿足an+1=an+n+a1,且a1=1,則1a1+1a2+1a28+1a29=.13.設(shè)數(shù)列an的前n項(xiàng)和為Sn,且Sn=4an-p,其中p是不為零的常數(shù).(1)求數(shù)列an的通項(xiàng)公式;(2)當(dāng)p=3時,數(shù)列bn滿足bn+1=bn+an,b1=2,求數(shù)列bn的通項(xiàng)公式.14.已知數(shù)列an滿足3Sn=(n+2)an,其中Sn為an的前n項(xiàng)和,a1=2.(1)求數(shù)列an的通項(xiàng)公式.(2)記數(shù)列1an的前n項(xiàng)和為Tn,是否存在無限集合M,使得當(dāng)nM時,總有|Tn-1|110成立?若存在,請找出一個這樣的集合;若不存在,請說明理由.15.已知數(shù)列an滿足a1=1256,an+1=2an,若bn=log2an-2,則b1b2bn的最大值為.16.我們把滿足xn+1=xn-f(xn)f(xn)的數(shù)列xn叫作牛頓數(shù)列.已知函數(shù)f(x)=x2-1,數(shù)列xn為牛頓數(shù)列,設(shè)數(shù)列an滿足an=lnxn-1xn+1,已知a1=2,則a3=.專題集訓(xùn)(三)1.C解析an-an-1=n(n2),a5=a1+(a2-a1)+(a3-a2)+(a4-a3)+(a5-a4)=1+2+3+4+5=15,故選C.2.C解析 由已知得a4=a2+a2=-12,a8=a4+a4=-24,a10=a8+a2=-30.3.C解析 利用遞推關(guān)系式可得,a2=3,a3=5,a4=9,a5=17,a6=33.故選C.4.C解析an+1=anan+1,1an+1=1an+1,又a1=12,1an是首項(xiàng)為2,公差為1的等差數(shù)列,1a8=2+71=9,即a8=19,故選C.5.21-12n解析 因?yàn)閍n+1=an+12n,所以an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=12n-1+12n-2+12+1=1-12n1-12=21-12n(n2),當(dāng)n=1時,a1=1也滿足上式,所以an=21-12n.6.B解析an+1-an=2n,an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2(n-1)+2(n-2)+2+36=n(n-1)+36(n2),當(dāng)n=1時,a1=36滿足上式,ann=n2-n+36n=n+36n-12n36n-1=11,當(dāng)且僅當(dāng)n=6時等號成立.故選B.7.C解析f(x)滿足f(n+1)=3f(n)+n3(nN*),f(n+1)-f(n)=n3,又f(1)=1,f(18)=f(1)+f(2)-f(1)+f(3)-f(2)+f(4)-f(3)+f(18)-f(17)=1+13+23+33+173=52,故選C.8.B解析 因?yàn)镾n=2an-1,所以當(dāng)n2時,Sn-1=2an-1-1,兩式相減得an=2an-2an-1,整理得an=2an-1,所以an是公比為2的等比數(shù)列.又因?yàn)閍1=2a1-1,所以a1=1,故數(shù)列an的通項(xiàng)公式為an=2n-1.由ann2,得2n-12n,所以n=1,2,3,4.故選B.9.C解析 由遞推關(guān)系式可得a2=23,a3=95.令bn=an+3n,則b1=5+3,b2=23+9,b3=95+27,若數(shù)列an+3n為等差數(shù)列,則b1+b3=2b2,即5+3+95+27=223+9,解得=-12.故選C.10.n2-n+33解析數(shù)列an滿足a1=33,an+1-an=2n,當(dāng)n2時,an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2(n-1)+2(n-2)+22+21+33=2(n-1)(n-1+1)2+33=n2-n+33,又a1=33滿足上式,故an=n2-n+33.11.1011解析an+1-an=2(n+1),an-an-1=2n(n2),an=(an-an-1)+(an-1-an-2)+(a3-a2)+(a2-a1)+a1=2n+2(n-1)+22+2=2n(n+1)2=n(n+1)(n2),又a1=2滿足上式,an=n(n+1),1an=1n(n+1)=1n-1n+1.設(shè)數(shù)列1an的前n項(xiàng)和為Sn,則S10=1-12+12-13+110-111=1-111=1011.12.2915解析 由an+1=an+n+a1且a1=1,得an+1=an+n+1,則an+1-an=n+1,所以an-an-1=n(n2),所以an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=n+(n-1)+2+1=n(n+1)2(n2),又a1=1滿足上式,所以an=n(n+1)2.因此1an=2n(n+1)=21n-1n+1,所以1a1+1a2+1a28+1a29=21-12+12-13+128-129+129-130=2915.13.解:(1)當(dāng)n=1時,S1=a1=4a1-p,所以a1=p3.因?yàn)镾n=4an-p,所以Sn-1=4an-1-p(n2),所以當(dāng)n2時,an=Sn-Sn-1=4an-4an-1,即3an=4an-1,所以anan-1=43,所以an=p343n-1.(2)當(dāng)p=3時,由(1)知,an=43n-1,由bn+1=bn+an,得bn+1-bn=43n-1,則bn=b1+(b2-b1)+(b3-b2)+(bn-bn-1)=2+1-(43)n-11-43=343n-1-1(n2),當(dāng)n=1時,b1=2也滿足上式,數(shù)列bn的通項(xiàng)公式為bn=343n-1-1.14.解:(1)由3Sn=(n+2)an,得3Sn-1=(n+1)an-1(n2),兩式相減得3an=(n+2)an-(n+1)an-1(n2),anan-1=n+1n-1(n2),an-1an-2=nn-2,a3a2=42,a2a1=31,又a1=2,由累乘法得an=n(n+1)(n2),當(dāng)n=1時,a1=2也滿足上式,an=n(n+1).(2)由(1)得,1an=1n(n+1)=1n-1n+1,Tn=1-12+12-13+13-14+1n-1n+1=nn+1.令|Tn-1|=nn+1-1=1n+19,故滿足條件的M存在,集合M=nN*|n9.15.6254解析 由題意可得log2an+1=log2(2an),即log2an+1=12log2an+1,整理可得(log2an+1-2)=12(log2an-2).又log2a1-2=-10,bn=log2an-2,所以數(shù)列bn是首項(xiàng)為-10,公比為12的等比數(shù)列,所以bn=-1012n-1=-522-n.令Sn=b1b2bn,則Sn=b1b2bn=(-5)n2n(3-n)2.當(dāng)n為偶數(shù)時,Sn可能取得最大值,由SnSn+2,SnSn-2(n=2k,kN*),可得n=4,則b1b2bn的最大值為6254.16.8解析f(x