2020版高考数学复习第五单元专题集训三由数列的递推关系式求通项公式练习理新人教A版.docx

专题集训三 由数列的递推关系式求通项公式 1.在数列an中,a1=1,an=2an-1+1(n2),则a5的值为()A.30B.31C.32D.332.已知数列an对任意的p,qN*都满足ap+q=ap+aq,且a2=-6,那么a10等于()A.-165B.-33C.-30D.-213.已知数列an的前n项和Sn对任意m,nN*都满足Sn+Sm=Sn+m,且a1=1,那么a10等于()A.1B.9C.10D.554.2018湖北部分重点中学二联 已知数列an满足a1=2,(2n-1)an+1=(2n+1)an(nN*),则a5=.5.2018吉林辽源田家炳中学等五校联考 数列an中,an+1=an1+3an,a1=2,则a10=.6.2018湖南长沙雅礼中学、河南省实验中学联考 在数列an中,a1=2,an+1n+1=ann+ln1+1n,则an=()A.2+nlnnB.2n+(n-1)lnnC.2n+nlnnD.1+n+nlnn7.2018湖南怀化模拟 在数列an中,已知a1=-14,an=1-1an-1(n2),则a2018的值为()A.2018B.-14C.45D.58.2018山东烟台模拟 已知an为等比数列,数列bn满足b1=2,b2=5,且an(bn+1-bn)=an+1,则数列bn的前n项和为()A.3n+1B.3n-1C.3n2+n2D.3n2-n29.2018哈尔滨六中月考 已知数列an满足a1=1,an+1an=12n,则a2017等于()A.121009B.122016C.122017D.12100810.2018安徽黄山检测 已知数列an满足a1=2,且an=2nan-1an-1+n-1(n2,nN*),则an=.11.2018湖南衡阳联考 已知数列an的前n项和为Sn,若Sn=2an-2n,则Sn=.12.已知数列an满足3Sn=(n+2)an(nN*),其中Sn为数列an的前n项和,a1=2.(1)求数列an的通项公式.(2)记数列1an的前n项和为Tn,是否存在无限集合M,使得当nM时,总有|Tn-1|110成立?若存在,请找出一个这样的集合;若不存在,请说明理由.13.2018山西榆社中学月考 设Sn为数列an的前n项和,2an-an-1=32n-1(n2),且3a1=2a2.记Tn为数列1an+Sn的前n项和,若对任意nN*,Tnm,则m的最小值为()A.13B.12C.23D.114.已知数列an的首项a1=1,其前n项和为Sn,且Sn+Sn+1=n2+2n+p,若an是递增数列,则实数p的取值范围是.专题集训(三)1.B解析a1=1,an=2an-1+1(n2),a2=2a1+1=3,a3=2a2+1=7,a4=2a3+1=15,a5=2a4+1=31.2.C解析 由已知得a4=a2+a2=-12,a8=a4+a4=-24,a10=a8+a2=-30.3.A解析a1=1,S1=1.Sn+Sm=Sn+m,令m=1,可得Sn+1=Sn+1,Sn+1-Sn=1,即当n1时,an+1=1,a10=1.4.18解析 由an+1an=2n+12n-1,得an=a1a2a1a3a2anan-1=231532n-12n-3=4n-2,则a5=18.5.255解析 数列an中,an+1=an1+3an,两边取倒数得1an+1=3+1an1an+1-1an=3,又1a1=12,1an=3n-52,a10=255.6.C解析 由题意得an+1n+1-ann=ln(n+1)-lnn,运用累加法得ann-a11=lnn-ln1=lnn,即ann=2+lnn,an=2n+nlnn,故选C.7.D解析a1=-14,an=1-1an-1(n2),a2=1-1-14=5,a3=1-15=45,a4=1-145=-14,数列an是周期数列,且周期为3,a2018=a6723+2=a2=5,故选D.8.C解析b1=2,b2=5,且an(bn+1-bn)=an+1,a1(b2-b1)=a2,即a2=3a1,又数列an为等比数列,数列an的公比q=3,bn+1-bn=an+1an=3,数列bn是首项为2,公差为3的等差数列,数列bn的前n项和Sn=2n+n(n-1)23=3n2+n2.故选C.9.D解析a1=1,an+1an=12n,an+1=12n1an,a2=121=12,a3=1222=12,a4=1232=122,a5=12422=122,a6=12522=123,a7=12623=123,a8=12723=124,a9=12824=124,以此类推,2017=1+10082,a2017=121008.故选D.10.n2n2n-1解析 由递推关系可得anan-1+(n-1)an=2nan-1(n2),则nan=12n-1an-1+12,即nan-1=12n-1an-1-1(n2),据此可得,数列nan-1是首项为1a1-1=-12,公比为12的等比数列,故nan-1=-1212n-1=-12n,则nan=1-12n=2n-12n,据此可得,数列an的通项公式为an=n2n2n-1.11.n2n解析Sn=2an-2n=2(Sn-Sn-1)-2n,整理得Sn-2Sn-1=2n,等式两边同时除以2n,有Sn2n-Sn-12n-1=1,又S1=2a1-2=a1,可得a1=S1=2,所以数列Sn2n是以1为首项,1为公差的等差数列,所以Sn2n=n,所以Sn=n2n.12.解:(1)由3Sn=(n+2)an得3Sn-1=(n+1)an-1(n2),两式相减得3an=(n+2)an-(n+1)an-1,anan-1=n+1n-1(n2),an-1an-2=nn-2,a3a2=42,a2a1=31,累乘得ana1=(n+1)n2,又a1=2,an=n(n+1).(2)由(1)知1an=1n(n+1)=1n-1n+1,Tn=1-12+12-13+13-14+1n-1n+1=nn+1,则|Tn-1|=nn+1-1=1n+1.由|Tn-1|110得1n+19,故满足条件的集合M存在,且集合M=n|n9,nN*.13.A解析 由2an-an-1=32n-1(n2),得an2n=14an-12n-1+34,an2n-1=14an-12n-1-1,由2an-an-1=32n-1(n2),可得2a2-a1=6,又3a1=2a2,2a1=6,解得a1=3.数列an2n-1是以12为首项,14为公比的等比数列,则an2n-1=1214n-1=122n-1,an=2n(21-2n+1)=21-n+2n,Sn=1+12+12n-1+(2+22+23+2n)=1-(12)n1-12+2(1-2n)1-2=22n-21-n,1an+Sn=121-n+2n+22n-21-n=132n,Tn=1312+122+12n=131-12n13.对任意nN*,Tnm,m的最小值为13.14.12,32解析 由Sn+Sn+1=n2+2n+p可得Sn-1+Sn=(n-1)2+2(n-1)+p(n2),两式相减得an+an+1=2n+1(n2),an-1+an=2n-1(n3),两式相减可得an+1-an-1=2(n3),数列a2,a4,a6,是以2为公差的等差数列,数列a3,a5,a7,是