初二数学 二次根式计算题
展开全部
a(n) = 1/[n^(1/2) + (n+1)^(1/2)]
= [(n+1)^(1/2) - n^(1/2)]/{[n^(1/2) + (n+1)^(1/2)] [(n+1)^(1/2) - n^(1/2)]}
= [(n+1)^(1/2) - n^(1/2)]/[(n+1) - n]
= (n+1)^(1/2) - n^(1/2)
a(1) + a(2) + ... + a(2010) + a(2011)
= 2^(1/2) - 1^(1/2) + 3^(1/2) - 2^(1/2) + ... + (2011)^(1/2) - (2010)^(1/2) + (2012)^(1/2) - (2011)^(1/2)
= (2012)^(1/2) - 1^(1/2)
= (2012)^(1/2) - 1.
{1/[1+2^(1/2)] + 1/[2^(1/2) + 3^(1/2)] + ... + 1/[(2011)^(1/2) + (2012)^(1/2)] } [1+(2012)^(1/2)]
= [(2012)^(1/2) - 1][ (2012)^(1/2) - 1]
= 2012 - 1
= 2011
= [(n+1)^(1/2) - n^(1/2)]/{[n^(1/2) + (n+1)^(1/2)] [(n+1)^(1/2) - n^(1/2)]}
= [(n+1)^(1/2) - n^(1/2)]/[(n+1) - n]
= (n+1)^(1/2) - n^(1/2)
a(1) + a(2) + ... + a(2010) + a(2011)
= 2^(1/2) - 1^(1/2) + 3^(1/2) - 2^(1/2) + ... + (2011)^(1/2) - (2010)^(1/2) + (2012)^(1/2) - (2011)^(1/2)
= (2012)^(1/2) - 1^(1/2)
= (2012)^(1/2) - 1.
{1/[1+2^(1/2)] + 1/[2^(1/2) + 3^(1/2)] + ... + 1/[(2011)^(1/2) + (2012)^(1/2)] } [1+(2012)^(1/2)]
= [(2012)^(1/2) - 1][ (2012)^(1/2) - 1]
= 2012 - 1
= 2011
本回答由网友推荐
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
|
广告 您可能关注的内容 |
