设数列{an}的前n项积为Tn,Tn=1-an,设cn=1/Tn(1)证明数列{Cn}是等差数列

2个回答

  • T1=a1=1-a1 2a1=1 a1=1/2

    a1a2...an=Tn=1-an (1)

    a1a2...a(n-1)=Tn-1=1-a(n-1) (2)

    (1)/(2)

    an=(1-an)/[1-a(n-1)]

    整理,得

    [2-a(n-1)]an=1

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

    a2=1/(2-1/2)=2/3

    假设当n=k(k∈N+,且k≥2)时,ak=k/(k+1),则当n=k+1时,

    a(k+1)=1/(2-ak)=1/[2-k/(k+1)]=1/[(2k+2-k)/(k+1)]=(k+1)/(k+2)=(k+1)/[(k+1)+1],同样满足.

    数列{an}的通项公式为an=n/(n+1)

    Tn=a1a2...an=(1/2)(2/3)...[n/(n+1)]=1/(n+1)

    cn=1/Tn=n+1

    c1=1/T1=1/a1=2

    cn-c(n-1)=(n+1)-n=1,为定值.

    数列{cn}是以2为首项,1为公差的等差数列.

    a1=1/2 am=m/(m+1) an=n/(n+1)

    假设存在满足题意的m、n.则n>2.

    2m/(m+1)=1/2 +n/(n+1)

    整理,得

    mn+3m=3n+1

    (n+3)m=(3n+1)

    m=(3n+1)/(n+3)=(3n+9-8)/(n+3)=3-8/(n+3)

    要m为大于1的正整数,0