已知数列{a}满足a1=a,对任意m,n属于N+,am+n=am*an恒成立

1个回答

  • 1.

    令m=1

    a(n+1)=an×a1=a×an

    题目没有给出a的取值范围,因此需要分类讨论:

    (1)

    a=0时,数列是各项均为0的常数数列,an=0

    (2)

    a≠0时,a(n+1)/an=a,为定值.

    数列是以a为首项,a为公比的等比数列.

    an=aⁿ

    数列{an}的通项公式为an=aⁿ.

    2.

    证:

    bn=anlnan=aⁿ×ln(aⁿ)=(nlna)×aⁿ

    b(n+1)/bn=[(n+1)lna]a^(n+1)/[(nlna)aⁿ]=a×(n+1)/n

    a>1 (n+1)/n>1

    b(n+1)/bn>1

    b(n+1)>bn

    数列{bn}是递增数列.

    3.

    证:

    cn=1/[loga(a(2n)×loga(a(2(n+1))]

    =1/[loga(a^(2n)) ×loga(a^(2(n+1))]

    =1/[2n×2(n+1)]

    =(1/4)[1/n(n+1)]

    =(1/4)[1/n -1/(n+1)]

    Sn=c1+c2+...+cn

    =(1/4)[1-1/2+1/2-1/3+...+1/n- 1/(n+1)]

    =(1/4)[1- 1/(n+1)]

    =1/4 -1/[4(n+1)]