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1微专题二数列通项公式的常用求法一、累加法、累乘法例1已知数列{an}满足an+1=an+2·3n+1,a1=3,则数列{an}的通项公式为________.答案an=3n+n-1解析由an+1=an+2·3n+1,得a2=a1+2×31+1,a3=a2+2×32+1,a4=a3+2×33+1,…,an=an-1+2×3n-1+1,累加可得an=a1+2×(31+32+…+3n-1)+(n-1),又a1=3,∴an=2·3-3n1-3+n+2=3n+n-1(n=1时也成立).例2设数列{an}是首项为1的正项数列,且(n+1)a2n+1-na2n+an+1an=0(n=1,2,3,…),则它的通项公式是an=________.答案1n解析原递推式可化为:[(n+1)an+1-nan](an+1+an)=0,∵an+1+an0,∴an+1an=nn+1,则a2a1=12,a3a2=23,a4a3=34,…,anan-1=n-1n,累乘可得ana1=1n,又a1=1,∴an=1n(n=1时也成立).跟踪训练1(1)在数列{an}中,a1=3,an+1=an+1nn+1,则数列{an}的通项公式为an=_______.答案4-1n解析原递推式可化为an+1=an+1n-1n+1,2则a2=a1+11-12,a3=a2+12-13,a4=a3+13-14,…,an=an-1+1n-1-1n,累加得an=a1+1-1n.故an=4-1n(n=1时也成立).(2)在数列{an}中,a1=1,an+1=2n·an,则an=______.答案(1)22nn解析a1=1,a2=2a1,a3=22a2,…,an=2n-1an-1,累乘得an=2·22·23·…·2n-1=(1)22nn,当n=1时也成立,故an=(1)22nn.二、换元法例3已知数列{an},其中a1=43,a2=139,且当n≥3时,an-an-1=13(an-1-an-2),求通项公式an.解设bn-1=an-an-1,原递推式可化为bn-1=13bn-2,{bn}是一个等比数列,b1=a2-a1=139-43=19,公比为13.∴bn=b1·13n-1=13n+1,故an-an-1=13n,an-1-an-2=13n-1,…,a3-a2=133,a2-a1=132,3用累加法得an=32-1213n,当n=1时也成立.跟踪训练2已知数列{an}中,a1=1,a2=2,当n≥3时,an-2an-1+an-2=1,求通项公式an.解当n≥3时,(an-an-1)-(an-1-an-2)=1,令bn-1=an-an-1,∴bn-1-bn-2=1,∴{bn}是等差数列,其中b1=a2-a1=1,公差为1,∴bn=n,∴b1+b2+…+bn-1=a2-a1+a3-a2+…+an-an-1=an-1,∴an-1=12n(n-1),∴an=n2-n+22(n=1时也成立).三、构造等差数列求通项例4已知数列{an}满足an+1=3an+2·3n+1,a1=3,求数列{an}的通项公式.解an+1=3an+2·3n+1,两边同除以3n+1,得an+13n+1=an3n+2,∴an3n是以a13=1为首项,以2为公差的等差数列,∴an3n=1+(n-1)×2=2n-1,∴an=(2n-1)·3n.例5若数列{an}中,a1=2且an=3+a2n-1(n≥2),求它的通项公式an.解将an=3+a2n-1两边平方整理,得a2n-a2n-1=3.数列{a2n}是以a21=4为首项,3为公差的等差数列.故a2n=a21+(n-1)×3=3n+1.因为an0,所以an=3n+1.例6已知数列{an}中,a1=1,且当n≥2时,an=an-12an-1+1,求通项公式an.解将an=an-12an-1+1两边取倒数,得1an-1an-1=2,4这说明1an是一个等差数列,首项是1a1=1,公差为2,所以1an=1+(n-1)×2=2n-1,即an=12n-1.跟踪训练3(1)已知数列{an}满足an+1=3an+3n,且a1=1.①证明:数列an3n是等差数列;②求数列{an}的通项公式.①证明由an+1=3an+3n,两边同时除以3n+1,得an+13n+1=an3n+13,即an+13n+1-an3n=13.由等差数列的定义知,数列an3n是以a13=13为首项,13为公差的等差数列.②解由(1)知an3n=13+(n-1)×13=n3,故an=n·3n-1,n∈N*.(2)已知数列{an}中,a1=1,an-1-an=anan-1(n≥2,n∈N*),则a10=________.答案110解析易知an≠0,∵数列{an}满足an-1-an=anan-1(n≥2,n∈N*),∴1an-1an-1=1(n≥2,n∈N*),故数列1an是等差数列,且公差为1,首项为1,∴1a10=1+9=10,∴a10=110.四、构造等比数列求通项例7已知数列{an}满足a1=1,an+1=3an+2,求数列{an}的通项公式.解由an+1=3an+2,可得an+1+1=3(an+1),又a1+1=2,∴{an+1}是以2为首项,以3为公比的等比数列,∴an+1=2·3n-1,∴an=2·3n-1-1.例8在数列{an}中,a1=-1,an+1=2an+4·3n-1,求通项公式an.解原递推式可化为an+1+λ·3n=2(an+λ·3n-1),①比较系数得λ=-4,①式为:an+1-4·3n=2(an-4·3n-1).则数列{an-4·3n-1}是一个等比数列,其首项为a1-4·31-1=-5,公比是2.∴an-4·3n-1=-5·2n-1,5即an=4·3n-1-5·2n-1.例9数列{an}满足a1=2,an+1=a2n(an0,n∈N*),则an=________.答案122n-解析因为数列{an}满足a1=2,an+1=a2n(an0,n∈N*),所以log2an+1=2log2an,即log2an+1log2an=2.又a1=2,所以log2a1=log22=1.故数列{log2an}是首项为1,公比为2的等比数列.所以log2an=2n-1,即an=122n-.跟踪训练4(1)若数列{an}中,a1=3且an+1=a2n(n是正整数),则它的通项公式是an=________.答案123n解析由题意知an0,将an+1=a2n两边取对数,得lgan+1=2lgan,即lgan+1lgan=2,又lga1=lg3,所以数列{lgan}是以lg3为首项,公比为2的等比数列,lgan=lga1·2n-1=12lg3n,故an=123n.(2)数列{an}中,a1=1,an+1=an3+4an,则an=________.答案13n-2解析由已知可得1an+1=3+4anan=3an+4,∴1an+1+2=3an+6=31an+2,又1a1+2=3,∴1an+2是以3为首项,以3为公比的等比数列,∴1an+2=3n,∴an=13n-2.(3)数列{an}中,已知a1=1,an+1=-2an+3n,则an=________.答案15·3n+25·(-2)n-1解析由已知可设an+1+λ·3n+1=-2(an+λ·3n),6比较系数可得λ=-15,即an+1-15·3n+1=-2an-15·3n,又a1-35=25,∴an-15·3n是以25为首项,-2为公比的等比数列,∴an-15·3n=25·(-2)n-1,∴an=15·3n+25·(-2)n-1.五、归纳推理法例10(1)设Sn是数列{an}的前n项和,且a1=-1,an+1=Sn+1Sn,则Sn=________.答案-1n解析由a1=-1,an+1=SnSn+1可得a2=S1S2=a1(a1+a2),故a2=12=11×2,同理可得a3=16=12×3,a4=112=13×4,…,由此猜想当n≥2时,有an=1n-1n=1n-1-1n,所以当n≥2时,Sn=a1+a2+…+an=-1+1-12+12-13+13-14+…+1n-1-1n=-1n.又因为S1=-1也适合上式,所以Sn=-1n.(2)已知数列{an}满足an+1=2an,0≤an12,2an-1,12≤an1.若a1=35,则a2018=________.答案15解析因为a1=35,根据题意得a2=15,a3=25,a4=45,a5=35,7所以数列{an}是以4为周期的数列,又2018=504×4+2,所以a2018=a2=15.跟踪训练5(1)在数列{an}中,an+1+(-1)nan=2n-1,则数列{an}的前12项和等于________.答案78解析由题意,当n为奇数时,an+1-an=2n-1,an+2+an+1=2n+1,两式相减得an+2+an=2;当n为偶数时,an+1+an=2n-1,an+2-an+1=2n+1,两式相加得an+2+an=4n.所以S12=(a1+a3+…+a11)+(a2+a4+…+a12)=2×3+4(2+6+10)=78.(2)已知数列{an}满足an+2=an+1-an,且a1=2,a2=3,Sn为数列{an}的前n项和,则S2018的值为________.答案5解析依题意得,a1=2,a2=3,a3=a2-a1=3-2=1,a4=a3-a2=1-3=-2,a5=a4-a3=-2-1=-3,a6=a5-a4=-3-(-2)=-1,a7=a6-a5=-1-(-3)=2,a8=a7-a6=2-(-1)=3,…,所以数列{an}是周期为6的周期数列.又因为2018=6×336+2,所以S2018=(2+3+1-2-3-1)×336+2+3=5.(3)用{x}表示不小于x的最小整数,例如{2}=2,{1.2}=2,{-1.1}=-1.已知数列{an}满足a1=1,an+1=a2n+an,则1a1+1+1a2+1+…+1a2018+1=________.答案1解析∵a1=1,an+1=a2n+an,∴an1,1an+1=1a2n+an=1anan+1=1an-1an+1,∴1an+1=1an-1an+1,8∴1a1+1+1a2+1+…+1a2018+1=1a1-1a2+1a2-1a3+…+1a2018-1a2019=1a1-1a2019=1-1a2019.∵01-1a20191,∴1-1a2019=1.
本文标题:(江苏专用)2020版高考数学大一轮复习 第六章 数列 微专题二 数列通项公式的常用求法教案(含解析
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