计算:(3^2+1)/(3^2-1)+ (5^2+1)/(5^2-1)+ (7^2+1)/(7^2-1)+…… +(99^2+1)/(99^2-
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解决时间 2021-02-19 14:13
- 提问者网友:我们很暧昧
- 2021-02-18 15:47
计算:(3^2+1)/(3^2-1)+ (5^2+1)/(5^2-1)+ (7^2+1)/(7^2-1)+…… +(99^2+1)/(99^2-1)
最佳答案
- 五星知识达人网友:鸠书
- 2021-02-18 16:19
设算是值为A
A-49=算式-49
=(3^2+1)/(3^2-1)-1+(5^2+1)/(5^2-1)-1+(7^2+1)/(7^2-1)-1+…… +(99^2+1)/(99^2-1)-1
=2/(3^2-1)+2/(5^2-1)+2/(7^2-1)+2/(99^2-1)=1/(3-1)-1/(3+1)+…… +1/(99-1)-1/(99+1)
=1/2-1/4+1/4-1/6+1/6-1/8+……1/98-1/100=1/2-1/100=49/100
A=49+49/100=49.49
大连卡耐基学校,网址www.dlbest.com.cn
A-49=算式-49
=(3^2+1)/(3^2-1)-1+(5^2+1)/(5^2-1)-1+(7^2+1)/(7^2-1)-1+…… +(99^2+1)/(99^2-1)-1
=2/(3^2-1)+2/(5^2-1)+2/(7^2-1)+2/(99^2-1)=1/(3-1)-1/(3+1)+…… +1/(99-1)-1/(99+1)
=1/2-1/4+1/4-1/6+1/6-1/8+……1/98-1/100=1/2-1/100=49/100
A=49+49/100=49.49
大连卡耐基学校,网址www.dlbest.com.cn
全部回答
- 1楼网友:时间的尘埃
- 2021-02-18 17:35
思路: [2n+1)^2+1]/[(2n+1)^2-1](n≥1) =1+1/(2n^2+2n) =1+1/2[1/n-1/(n+1)] (3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+…+(2005^2+1)/(2005^2-1) =1+1/2*(1-1/2)+1+1/2*(1/2-1/3)…+1+1/2*(1/1001-1/1002)+1+1/2*(1/1002-1/1003) =1*1002+1/2[(1-1/2)+(1/2-1/3)+...+(1/1001-1/1002)+(1/1002-1/1003)] =1002+1/2[1-1/1003] =1002+501/1003
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