求和1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+.+1/(n^2-1)具体过程
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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](/uploads/image/z/1432019-11-9.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)
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[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)]