Question

求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+......+1/(n^2-1)具体过程

帮忙写具体过程，谢谢！

Answer 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)]

Answer 2

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)]

Answer 3

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)]