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

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

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(n≥2,n∈N*),则a5=(　　)A.6B.10C.15D.212.已知数列{an}对任意的p,q∈N*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(n∈N*),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ann≤2的正整数n的集合为(　　)A.{1,2}B.{1,2,3,4}C.{1,2,3}D.{1,2,4}9.[2018·南昌模拟]已知数列{an}满足an=3an-1+3n-1(n≥2),且a1=5.若an+λ3n为等差数列,则实数λ=(　　)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n∈N*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.n14.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n∈M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b1·b2·…·bn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(n≥2),∴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+7×1=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(n≥2),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(n≥2),当n=1时,a1=36满足上式,∴ann=n2-n+36n=n+36n-1≥2n·36n-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(n∈N*),∴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.n8.B　[解析]因为Sn=2an-1,所以当n≥2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ann≤2,得2n-1≤2n,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+λ3n,则b1=5+λ3,b2=23+λ9,b3=95+λ27,若数列an+λ3n为等差数列,则b1+b3=2b2,即5+λ3+95+λ27=2×23+λ9,解得λ=-12.故选C.10.n2-n+33　[解析]∵数列{an}满足a1=33,an+1-an=2n,∴当n≥2时,an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2(n-1)+2(n-2)+…+2×2+2×1+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(n≥2),∴an=(an-an-1)+(an-1-an-2)+…+(a3-a2)+(a2-a1)+a1=2n+2(n-1)+…+2×2+2=2×n(n+1)2=n(n+1)(n≥2),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(n≥2),所以an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=n(n+1)2(n≥2),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=2×1-12+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(n≥2),所以当n≥2时,an=Sn-Sn-1=4an-4an-1,即3an=4an-1,所以anan-1=43,所以an=p3·43n-1.(2)当p=3时,由(1)知,an=43n-1,由bn+1=bn+an,得bn+1-bn=43n-1,n则bn=b1+(b2-b1)+(b3-b2)+…+(bn-bn-1)=2+1-(43) n-11-43=3·43n-1-1(n≥2),当n=1时,b1=2也满足上式,∴数列{bn}的通项公式为bn=3·43n-1-1.14.解:(1)由3Sn=(n+2)an,得3Sn-1=(n+1)an-1(n≥2),两式相减得3an=(n+2)an-(n+1)an-1(n≥2),∴anan-1=n+1n-1(n≥2),∴an-1an-2=nn-2,…,a3a2=42,a2a1=31,又a1=2,由累乘法得an=n(n+1)(n≥2),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+1<110,得n>9,故满足条件的M存在,集合M={n∈N*|n>9}.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=-10×12n-1=-5×22-n.令Sn=b1·b2·…·bn,则Sn=b1·b2·…·bn=(-5)n×2n(3-n)2.当n为偶数时,Sn可能取得最大值,由Sn≥Sn+2,Sn≥Sn-2(n=2k,k∈N*),可得n=4,则b1·b2·…·bn的最大值为6254.16.8　[解析]∵f(x)=x2-1,∴f'(x)=2x.又∵数列{xn}为牛顿数列,∴xn+1=xn-f(xn)f'(xn)=xn-xn2-12xn=12xn+1xn,∴an+1=lnxn+1-1xn+1+1=ln12(xn+1xn)-112(xn+1xn)+1=ln(xn-1)2(xn+1)2=2lnxn-1xn+1=2an.又a1=2,∴数列{an}是以2为首项,2为公比的等比数列,∴a3=2×22=8.