高中数学第二章数列2-5等比数列的前n项和第3课时数列的通项公式优化练习新人教A版必修52019-2020学年度

an+1n所以=, ann+1故an=anan-1a2n-1n-2122··…··a1=··…··=. an-1an-2a1nn-1233n[B组 能力提升] 11.已知数列{an}满足a1=,a1+a2+…+an=n2an,则an为( ) 2A.an=C.an=1n n+1 B.an=D.an=1n-1 n+1 解析:∵a1+a2+…+an=n2an,① ∴a1+a2+…+an-1= (n-1)2an-1(n≥2,n∈N*),② ①-②得an=n2an-(n-1)2an-1. ann-1即=(n≥2,n∈N*). an-1n+1a2a3a4an1234n-2n-1∴···…·=××××…××. a1a2a3an-13456nn+1an即=a121,又a1=,∴an=211=成立, 21, 当n=1时,a1=∴an=答案:A 1(n∈N*). 2.已知{an}是首项为1的正项数列,且(n+1)a2n+1-na2n+anan+1=0,则{an}的通项公式为an=( ) 1A. n1C. n+1B.(D.(nn-1) n+1nn) n+1解析:∵(n+1)a2n+1-na2n+anan+1=0. ∴(an+1+an)·[(n+1)an+1-nan]=0. 5 / 7 ∵an>0,∴an+1+an>0. an+1nn∴=,即an+1=an. ann+1n+1∴an=n-1n-1n-2n-1n-2n-3211an-1=·an-2=…=···…···a1=nnn-1nn-1n-232n(n≥2). 11当n=1时,a1=也成立,∴an=. nn答案:A 3.对于数列{an},满足a1=1,an+1=an+解析:∵an+1-an=n+1-n, ∴(a2-a1)+(a3-a2)+…+(an-an-1)=(2-1)+(3-2)+…+(n-n-1),即an=n(n≥2),将n=1代入也成立,∴an=n. 答案:n 4.设数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2)(n∈N*),则通项an=________. 解析:数列{nan}的前n项和为a1+2a2+3a3+…+nan=n(n+1)(n+2).① 其前n-1项和为a1+2a2+3a3+…+(n-1)an-1=(n-1)n(n+1).② ①-②,得nan=n(n+1)[(n+2)-(n-1)]=3n(n+1),即an=3n+3. 当n=1时也满足上式.故an=3n+3. 答案:3n+3 5.已知数列{an}满足a1=1,an+1=2an+1. (1)证明数列{an+1}是等比数列; (2)求数列{an}的通项公式. 解析:(1)证明:法一:因为an+1=2an+1, 所以an+1+1=2(an+1). 由a1=1,知a1+1≠0,从而an+1≠0. an+1+1所以=2(n∈N*). an+1所以数列{an+1}是等比数列. 6 / 7 1,则an=________. n+1+n 法二:由a1=1,知a1+1≠0,从而an+1≠0. an+1+12an+1+1∵==an+1an+1∴{an+1}是等比数列. (2)由(1)可知an+1=2×2n-1=2n,∴an=2n-1. 6.数列{an}的前n项和为Sn,a1=1,Sn+1=4an+2(n∈N*). (1)设bn=an+1-2an,求证:{bn}是等比数列; (2)设cn=an,求证:{cn}是等比数列. 3n-1an+1=2(n∈N*), 证明:(1)由Sn+1=4an+2得Sn=4an-1+2,an+1=Sn+1-Sn=(4an+2)-(4an-1+2)=4an-4an-1(n≥2), 即an+1-2an=2(an-2an-1), ∴bn=2bn-1(n≥2,n∈N*),又b1=a2-2a1=3, ∴{bn}是以3为首项,2为公比的等比数列. (2)由(1)知an+1-2an=bn=3·2n-1,于是有 an-21an-1=3·2n-2, 21an-1-22an-2=3·2n-2, 22an-2-23an-3=3·2n-2, … 2n-2a2-2n-1a1=3·2n-2. 将以上n-1个等式叠加得 an-2n-1a1=(n-1)·3·2n-2, ∴an=3(n-1)2n-2+2n-1a1=(3n-1)·2n-2(n≥2,n∈N*), an又n=1时也满足此式,∴cn==2n-2, 3n-1∴{cn}是等比数列,公比是2. 7 / 7

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