1+b+b
∵|b|>1, ∴ lim n→∞ 1+b+b2+…+bn-1 bn= lim n→∞
1×(1-bn) 1-b bn= lim n→∞
1 bn-1 1-b= 1 b-1. 答案: 1 b-1.
已知数列bn,满足b1=1,b2=5,bn+1=5bn-6bn-1(n≥2),若数列an满足a1=1,an=bn(1/b
已知数列bn满足bn=b^2n,其前n项和为Tn,求(1-bn)/Tn
已知数列an,bn,cn满足[a(n+1)-an][b(n+1)-bn]=cn
18、一道数列题已求出数列An=2n.若数列Bn满足B(n+1)=Bn^2-(n-2)Bn+3,Bn大于等于1,证明:B
己知an是等差数列 a2+a4=6 a5=5 {bn}满足bn=anan+1则1/b1+1/b2+……+1/bn等于
已知等比数列{an}中,a1=3,a4=81,若数列{bn}满足bn=log3an,则数列{1b
已知a b 是常数 lim(a根号(2n^2+n+1) -bn))=1 则a+b=
等差数列a3=2 a8=12,满足{bn} b1=4 an+b(n-1)=bn 求通项公式
已知数列{an}{bn}满足a1=1,a2=3,b(n+1)/bn=2,bn=a(n+1)-an,(n∈正整数),求数列
已知数列an,bn满足a1=1,a2=3,(b(n)+1)/bn=2,bn=a(n+1)-an,(n∈正整数)
设a1=2,a2=4,数列{bn}满足:bn=a(n+1)-an,b(n+1)=2bn+2.
急 设A1=2,A2=4,数列Bn满足:Bn=A(n+1)-An,B(n+1)=2Bn +2
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