求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(n^2-1)具体过程
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/21 03:07:12
![求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(n^2-1)具体过程](/uploads/image/z/6843621-21-1.jpg?t=%E6%B1%82%E5%92%8C1%2F%282%5E2-1%29%2B1%2F%283%5E2-1%29%2B1%2F%284%5E2-1%29%2B.%2B1%2F%28n%5E2-1%29%E5%85%B7%E4%BD%93%E8%BF%87%E7%A8%8B)
求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(n^2-1)具体过程
求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(n^2-1)具体过程
求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(n^2-1)具体过程
1/(2-1)(2+1)+1/(3-1)(3+1)+1/(4-1)(4+1)+...+1/(n-1)(n+1)
=1/2[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-1)-1/(n+1)]
=1/2[1-1/(n+1)]
=n/[2(n+1)]
1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+......+1/(n^2-1)
=1/(2-1)(2+1)+1/(3-1)(3+1)+1/(4-1)(4+1)+......+1/(n-1)(n+1)
=1/(1*3)+1/(2*4)+1/(3*5)+……+1/(n-1)(n+1)
=1/2*(1-1/3+1/2-1/4+1/3-1/5+……+1/(n-1)-1/(n+1))
=1/2(1+1/2-1/n-1/(n+1))
=(3n+2)(n-1)/[4n(n+1)]
1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+......+1/(n^2-1)
=1/[(2-1)(2+1)]+1/[(3-1)(3+1)]+....+1/[(n-1)(n+1)]
=1/(1*3)+1/(2*4)+1/(3*5)+1/(4*6)+.....+1/[(n-1)(n+1)]
=1/2(1-1/3)+1/2(1/2-1/4)+1/2(1/3-1/5...
全部展开
1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+......+1/(n^2-1)
=1/[(2-1)(2+1)]+1/[(3-1)(3+1)]+....+1/[(n-1)(n+1)]
=1/(1*3)+1/(2*4)+1/(3*5)+1/(4*6)+.....+1/[(n-1)(n+1)]
=1/2(1-1/3)+1/2(1/2-1/4)+1/2(1/3-1/5)+1/2(1/4-1/6)+....+1/2[1/(n-1)-1/(n+1)]
=1/2[1-1/3+1/2+1/4+1/3-1/5+1/4-1/6+...+1/(n-1)-(n+1)]
=1/2[1+1/2-1/n-1/(n+1)]
=(3n+2)(n-1)/[4n(n+1)]
收起