2020版高考数学复习第五单元专题探究3由数列的递推关系式求通项公式练习.docx

专题探究3由数列的递推关系式求通项公式 1.2018沈阳联考 已知数列an满足a1=1,an=an-1+n(n2,nN*),则a5=()A.6B.10C.15D.212.已知数列an对任意的p,qN*满足ap+q=ap+aq,且a2=-6,则a10等于()A.-165B.-33C.-30D.-213.已知数列an满足a1=2,an+1=2an-1,则a6=()A.31B.32C.33D.344.2018郑州模拟 已知数列an满足an+1=anan+1,且a1=12,则a8=()A.17B.18C.19D.1105.2018河南师大附中模拟 已知数列an满足an+1=an+12n,且a1=1,则an=.6.已知数列an满足a1=36,an+1=an+2n,则ann的最小值为()A.10B.11C.12D.137.2018中山模拟 设函数f(x)满足f(n+1)=3f(n)+n3(nN*),且f(1)=1,则f(18)=()A.20B.38C.52D.358.已知数列an的前n项和Sn=2an-1,则满足ann2的正整数n的集合为()A.1,2B.1,2,3,4C.1,2,3D.1,2,49.2018南昌模拟 已知数列an满足an=3an-1+3n-1(n2),且a1=5.若an+3n为等差数列,则实数=()A.2B.5C.-12D.1210.2018铜仁模拟 已知数列an满足a1=33,an+1-an=2n,则an=.11.2018重庆江津中学、合川中学等七校联考 在数列an中,a1=2,an+1-an=2(n+1),则数列1an的前10项的和为.12.2018郑州联考 设数列an对任意nN*都满足an+1=an+n+a1,且a1=1,则1a1+1a2+1a28+1a29=.13.设数列an的前n项和为Sn,且Sn=4an-p,其中p是不为零的常数.(1)求数列an的通项公式;(2)当p=3时,数列bn满足bn+1=bn+an,b1=2,求数列bn的通项公式.14.已知数列an满足3Sn=(n+2)an,其中Sn为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.15.已知数列an满足a1=1256,an+1=2an,若bn=log2an-2,则b1b2bn的最大值为.16.我们把满足xn+1=xn-f(xn)f(xn)的数列xn叫作牛顿数列.已知函数f(x)=x2-1,数列xn为牛顿数列,设数列an满足an=lnxn-1xn+1,已知a1=2,则a3=.专题集训(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,故选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 利用递推关系式可得,a2=3,a3=5,a4=9,a5=17,a6=33.故选C.4.C解析an+1=anan+1,1an+1=1an+1,又a1=12,1an是首项为2,公差为1的等差数列,1a8=2+71=9,即a8=19,故选C.5.21-12n解析 因为an+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),当n=1时,a1=1也满足上式,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),当n=1时,a1=36满足上式,ann=n2-n+36n=n+36n-12n36n-1=11,当且仅当n=6时等号成立.故选B.7.C解析f(x)满足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,故选C.8.B解析 因为Sn=2an-1,所以当n2时,Sn-1=2an-1-1,两式相减得an=2an-2an-1,整理得an=2an-1,所以an是公比为2的等比数列.又因为a1=2a1-1,所以a1=1,故数列an的通项公式为an=2n-1.由ann2,得2n-12n,所以n=1,2,3,4.故选B.9.C解析 由递推关系式可得a2=23,a3=95.令bn=an+3n,则b1=5+3,b2=23+9,b3=95+27,若数列an+3n为等差数列,则b1+b3=2b2,即5+3+95+27=223+9,解得=-12.故选C.10.n2-n+33解析数列an满足a1=33,an+1-an=2n,当n2时,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满足上式,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满足上式,an=n(n+1),1an=1n(n+1)=1n-1n+1.设数列1an的前n项和为Sn,则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,则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满足上式,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)当n=1时,S1=a1=4a1-p,所以a1=p3.因为Sn=4an-p,所以Sn-1=4an-1-p(n2),所以当n2时,an=Sn-Sn-1=4an-4an-1,即3an=4an-1,所以anan-1=43,所以an=p343n-1.(2)当p=3时,由(1)知,an=43n-1,由bn+1=bn+an,得bn+1-bn=43n-1,则bn=b1+(b2-b1)+(b3-b2)+(bn-bn-1)=2+1-(43)n-11-43=343n-1-1(n2),当n=1时,b1=2也满足上式,数列bn的通项公式为bn=343n-1-1.14.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(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),当n=1时,a1=2也满足上式,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,故满足条件的M存在,集合M=nN*|n9.15.6254解析 由题意可得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,所以数列bn是首项为-10,公比为12的等比数列,所以bn=-1012n-1=-522-n.令Sn=b1b2bn,则Sn=b1b2bn=(-5)n2n(3-n)2.当n为偶数时,Sn可能取得最大值,由SnSn+2,SnSn-2(n=2k,kN*),可得n=4,则b1b2bn的最大值为6254.16.8解析f(x