数学计算题？
3个回答
展开全部
an
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1011 又 2025分之1012
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1011 又 2025分之1012
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
展开全部
an
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1012 又 2025分之1012
= 4n^2/[(2n-1)(2n+1)]
=1 + 1/[(2n-1)(2n+1)]
=1 + (1/2)[ 1/(2n-1) -1/(2n+1) ]
Sn
=a1+a2+...+an
=n + (1/2)[ 1 -1/(2n+1) ]
=n+ n/(2n+1)
(2x2)/(1x3)+ (4x4)/(3x5)+...+(2024x2024)/(2023x2025)
=S1012
=1012 + 1012/(2025)
=1012 又 2025分之1012
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 |
