an=a^(5-n),bn=n k,c5

由bn=an-1与an-1=an[(an+1)-1]得bn=[bn+1]*（bn+1)所以bn/[bn+1]=(bn+1)所以[bn+1]/bn=1/(bn+1)即1/bn+1=(bn+1)所以{1/

A(n+1)=(1+1/n)An+(n+1)/2^nA(n+1)=(n+1)/n×An+(n+1)/2^n两边除n+1A(n+1)/(n+1)=An/n+1/2^nB(n+1)=Bn+1/2^nBn=

(n+1)/bn=2∴bn=b1×2^(n-1)b1=a2-a1=3-1=2∴bn=2^n∴a(n+1)-an=2^n∴a2-a1=2a3-a2=2^2a4-a3=2^3……an-a(n-1)=2^(

你应该是题目打错了,(b(n)+1)/bn=2,这个条件应该是b(n+1)/bn=2吧因为如果是你所说的bn将恒等于1等于1不要紧,关键是这样的话b1=a2-a1=2且b1=1矛盾如果是我所说条件的话

当an,bn各取前9项时a1+a2+a3+...+a9/b1+b2+b3+...+b9=7*9+2/9+3.=65/12a5,b5是等差中项a5/b5=a1+a2+a3+...+a9/b1+b2+b3

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

(1)an*a(n-1)+1=2a(n-1)an=[2a(n-1)-1]/a(n-1)an-1=[2a(n-1)-1]/a(n-1)-1=[a(n-1)-1]/a(n-1)1/(an-1)=a(n-1

A(n+1)=2An+KA(n)=2A(n-1)+KA(n+1)-An=2[An-A(n-1)]Bn=A(n+1)-AnBn-1=An-A(n-1)Bn=2B(n-1){Bn}为等比数列

(N+1)是下标么?5对什么,看不太懂

(1)an*a(n-1)+1=2a(n-1)an=[2a(n-1)-1]/a(n-1)an-1=[2a(n-1)-1]/a(n-1)-1=[a(n-1)-1]/a(n-1)1/(an-1)=a(n-1

1.a(n+1)=2an-a(n-1)a(n+1)-an=an-a(n-1)an为以1/4为首项,1/2为公差的等差数列an=n/2-1/4bn-an=bn-n/2+1/4b(n+1)-a(n+1)=

首先a＝0,否则极限不存在.又lim(n→∞)[(an^2+bn+c)/(2n+5)]＝lim(n→∞)[(bn+c)/(2n+5)]＝lim(n→∞)[(b+c/n)/(2+5/n)]＝b/2=3∴

a1+a2+...+an=a*n^2+bnan=4n-5/2,易知｛an｝为等差数列利用等差数列求和公式得：n[3/2+4n-（5/2）]/2=a*n^2+bnn(4n-1)=2a*n^2+2bn4n

(1)bn=a(n+1)-1/2an,b(n+1)=a(n+2)-1/2a(n+1)sob(n+1)/bn=……将b(n+1)和bn中的a(n+2)a(n+1)和an全部化为an,可得b(n+1)/b

证明：a(n+2)=[an+a(n+1)]/2a(n+2)-a(n+1)=-[a(n+1)-an]/2,即b(n+1)=-bn/2,b(n+1)/bn=-1/2,b1=a2-a1=1-0=1所以bn是

2a(n+1)-an=n-2/n(n+1)(n+2)2a(n+1)-2/(n+1)(n+2)=an-1/n(n+1)[a(n+1)-1/(n+1)(n+2)]/[an-1/n(n+1)]=1/2bn=

设an=a1+(n-1)d,bn=an+a(n-1)=a1+(n-1)d+a1+nd=2a1+(2n-1)dbn为首项为2a1-d,公差为2d的等差数列

1、证明：a1=λ,a2=(2/3)a1+1-4=2λ/3-3,a3=(2/3)a2+2-4=4λ/9-4.若λ=0,a1=0,显然{an}不是等比数列；若λ≠0,则a2/a1=2/3-3/λ,a3/

(1)a(n+1)-an=(n+1+2013)-(n+2013)=1∴b(n+1)-bn=cn/[a(n+1)-an]=cn=2^n+n∴bn-b(n-1)=2^(n-1)+n-1...b2-b1=2

根据数列求和公式Sn=（a1+an)*n/2An/Bn=[(a1+an)*n/2]/[(b1+bn)*n/2]=(a1+an)/(b1+bn)由等差数列有a1+an=2*a[(1+n)/2]这里方括号