解析：
1/[n(n+2)]=(1/2)[(n+2)-n]/[n(n+2)]=(1/2)[1/n-1/(n+2)]
∴原式
=(1/2)*[1-1/3+1/3-1/5+1/5-1/7+…………+1/97-1/99]
=(1/2)*[1-1/99]
=(1/2)*98/99
=49/99
不懂欢迎再问
1/1*3+1/3*5+1/5*7+...1/97*99
1个回答
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