设等比数列an的首项a1>1,公比q>1,求证:数列{loganan+1】是递减数列
问题描述:
设等比数列an的首项a1>1,公比q>1,求证:数列{loganan+1】是递减数列
这里所有的n、n+1都是下角标,那个log是以an为底,a(n+1)是直接跟在log后面
答
loganan+1 -log(an-1)an=logan(an×q)-log(an-1)(an-1×q)
=1+loganq-1-log(an-1)q
=loganq-log(an-1)q<0
所以递减