已知数列{an},an≥0，a1=0,（an+1）^2+a（n+1）-1=an^2(n∈N+)求证当n∈N+时，an<an+1
3个回答
展开全部
First, we should prove that an<1.
Because a1=0<1, if 0<=an<1, then (an+1)^2+a(n+1)<2, then 0<=an+1<1.
Second, we prove the conclusion.
Because 0<=an<1 for all positive integer n, so a(n+1)-1<0, then (an+1)^2>an^2.
So an<an+1.
Because a1=0<1, if 0<=an<1, then (an+1)^2+a(n+1)<2, then 0<=an+1<1.
Second, we prove the conclusion.
Because 0<=an<1 for all positive integer n, so a(n+1)-1<0, then (an+1)^2>an^2.
So an<an+1.
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
