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.求通项公式an2.若a>1,bn=anInan,求证：数列{bn}为递增数列3.若a>o,a不等于1,cn=1/logaa2nlogaa2(n+1),求证：数列{cn}倩n项和Sn

Question

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.求通项公式an
2.若a>1,bn=anInan,求证：数列{bn}为递增数列
3.若a>o,a不等于1,cn=1/logaa2nlogaa2(n+1),求证：数列{cn}倩n项和Sn

Answer 1

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)]