已知,n >0,化简根号下1+1/n²+1/(n+1)²
1个回答
展开全部
1+1/n^2+1/(n+1)^2
={[n(n+1)]^2 +(n+1)^2 +n^2}/ [n(n+1)]^2
={(n^2 +1)*(n+1)^2 +n^2}/ [n(n+1)]^2
={(n^2 +1)*(n^2 +2n+1)^2 +n^2}/ [n(n+1)]^2
令 n^2 +1=a
则 原式可化为
={a*(a+2n)+n^2} / [n(n+1)]^2
={a^2 +2an +n^2} / [n(n+1)]^2
={a+n}^2 / [n(n+1)]^2
所以 根号[1 +1/n^2 +1/(n+1)^2]
=(a+n)/n(n+1)
=(n^2 +n+1)/n(n+1)
={[n(n+1)]^2 +(n+1)^2 +n^2}/ [n(n+1)]^2
={(n^2 +1)*(n+1)^2 +n^2}/ [n(n+1)]^2
={(n^2 +1)*(n^2 +2n+1)^2 +n^2}/ [n(n+1)]^2
令 n^2 +1=a
则 原式可化为
={a*(a+2n)+n^2} / [n(n+1)]^2
={a^2 +2an +n^2} / [n(n+1)]^2
={a+n}^2 / [n(n+1)]^2
所以 根号[1 +1/n^2 +1/(n+1)^2]
=(a+n)/n(n+1)
=(n^2 +n+1)/n(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