问题标题：
初一有理数计算题,1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+...+1/(19^2-1)+1/(20^2-1)
问题描述：
初一有理数计算题,
1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+...+1/(19^2-1)+1/(20^2-1)
付津生回答：
1/(2^2-1)+1/(3^2-1)+1/(4^2-1)+...+1/(19^2-1)+1/(20^2-1)
=1/(2^2-1^2)+1/(3^2-1^2)+1/(4^2-1^2)+...+1/(19^2-1^2)+1/(20^2-1^2)
=1/(2+1)(2-1)+1/(3+1)(3-1)+……+1/(20+1)(20-1)
=1/2*3+1/3*4+...+1/20*21
=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)...+(1/19-1/20)
=1/2-1/20
=9/20
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