设n为自然数,在数1/n和n+1之间插入n个正数,使n+2个正数成等比数列
1个回答
展开全部
设公比为q
a1 = 1/n
a(n+2) = n+1 = 1/n * q^(n+1)
q = [n(n+1)] ^ (1/(n+1))
a1 = 1/n
a2 = (1/n) * [n(n+1)] ^ (1/(n+1))
a3 = (1/n) * [n(n+1)] ^ (2/(n+1))
.
a(n+2) = (1/n) * [n(n+1)] ^ ((n+1)/(n+1)) =n+1
a1 = 1/n
a(n+2) = n+1 = 1/n * q^(n+1)
q = [n(n+1)] ^ (1/(n+1))
a1 = 1/n
a2 = (1/n) * [n(n+1)] ^ (1/(n+1))
a3 = (1/n) * [n(n+1)] ^ (2/(n+1))
.
a(n+2) = (1/n) * [n(n+1)] ^ ((n+1)/(n+1)) =n+1
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