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)]
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
数学人气：163 ℃时间：2020-06-16 04:24:46
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