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已知数列{An},满足A1=1,An+1=2nAn,求数列{An}通...

由an+1=2nan,得an+1an=2n,∴n≥2时,anan?1=2n-1,∴n≥2时,an=a1×a2a1×a3a2×…×anan?1=1×2×22×…×2n-1=21+2+…+(n-1)=2n(n?1)2,又a1=1适合上式,∴an=2n(n?1)2.

解(Ⅰ)a2=5,a3=7,a4=9,猜想an=2n+1.…(4分)(Ⅱ)Sn=n(3+2n+1)2=n2+2n,…(6分)使得2n>Sn成立的最小正整数n=6.…(7分)下面给出证明:n≥6(n∈N*)时都有2n>n2+2n.①n=6时,26>62+2×6,即64>48成立;…(8分)②假设n=k(k≥6,k∈N*)时...

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Sn = (n+1)/2 a(n+1) ——① S(n-1) = n/2 an ——② ①-② an = [n+1]/2 a[n+1] - n/2 an [n+2]/2 an = [n+1]/2 a[n+1] a[n+1] / an =[ n+2] /[ n+1] 用累乘法: a2 / a1 = 3/2 a3 / a2 = 4/3 a4 / a3 = 5/4 ︰ an / a[n-1] =[ n+1 ]/ n ∴ an / a1 = ...

(1)由a1=2,an+1=n+2nan,(n∈N*).所以a2=1+21×2=6,…(2分)同理a3=2+22×6=12,a4=3+23×12=20…(4分)(2)猜想an=n(n+1)…(6分)证明:①当n=1时,猜想成立.…(7分)②设当n=k时(n∈N*)时,猜想成立,即ak=k(k+1),…(8分)则当n=k+1...

(1)∵a1=2,an+1=an2-nan+1∴a2=a12-a1+1=3a3=a22-2a2+1=4a4=a32-3a3+1=5故猜想an=n+1.(2)用数学归纳法证明:①当n=1时,a1≥3=1+2,不等式成立.②假设当n=k时不等式成立,即ak≥k+2,那么ak+1=ak(ak-k)+1≥(k+2)(k+2-k)+1=2k+5≥k+3.也就...

1. n≥2时, a1+2a2+3a3+...+nan=[(n+1)/2]a(n+1) (1) a1+2a2+3a3+...+(n-1)a(n-1)=(n/2)an (2) (1)-(2) nan=[(n+1)/2]a(n+1)-(n/2)an (n+1)a(n+1)=3nan [(n+1)a(n+1)]/(nan)=3,为定值 a1×1=1×1=1,数列{nan}是以1为首项,3为公比的等比数列 nan...

(1)∵an+1=3an+4,∴an+1+2=3(an+2),…(3分)又∵a1=1,∴a1+2=3≠0,∴数列{an+2}是以3为首项,3为公比的等比数列 …(6分)(2)由(1)得 an+2=3×3n-1=3n∴an=3n-2,bn=n?an=n?3n-2n …(8分)∴Tn=(1×31-2×1)+(2×32-2×2)+…+(n?3n-2n)=1×31+...

令n=1,由a1=2及na(n+1)=Sn+n(n+1)①得a2=4,故a2-a1=2,当n≥2时,有(n-1)an=S(n-1)+n(n-1)②①-②得:na(n+1)-(n-1)an=an+2n整理得,a(n+1)-an=2(n≥2)当n=1时,a2-a1=2,所以数列{an}是以2为首项,以2为公差的等差数列,故an=2n ...

nan+1=(n+1)an两边同除以n(n+1), 得 a(n+1)/(n+1)=an/n 令bn=an/n 则b(n+1)=a(n+1)/(n+1) ∴b(n+1)-bn=0 b1=a1/1=2 所以数列{bn}是首项为2公差为0的等差数列 由等差数列公式 bn=2 你题目抄错了! 应该是 在数列{an}中,a1=2,na(n+1)-1=(n+...

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