1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n*(n+1))的计算公式

1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n*(n+1))的计算公式
1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n*(n+1))的计算公式

1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n*(n+1))的计算公式
1/1*2+1/2*3+1/3*4+1/4*5+...+1/n*[n+1]
=1/1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/n+1
=1/1-1/n+1
=n/(n+1)

1-1/2+1/2-1/3+...
总之1/(a*(a+1))=1/a-1/a+1 这是常用的裂项公式

1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n*(n+1))
=1-1/2+1/2-1/3……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
有不懂欢迎追问

这是列项相消法
1/(1*2)+1/(2*3)+1/(3*4)+…+1/(n*(n+1))
=1-1/2+1/2-1/3……+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
有不懂欢迎追问

将式子的每一项拆开，拆成(1-1/2)+(1/2-1/3)+....+(1/n-1/(n+1))=1-(1/(n+1))