1/1*2+1/2*3+1/3*4+。。。+1/(n(n+1))=?

来源：百度知道 编辑：UC知道 时间：2024/05/31 18:04:41
数列。。。

1/1*2+1/2*3+1/3*4+...+1/n(n+1)

=(1/1)-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+...+(1/n)-1/(n+1)

=1-1/(n+1)

=(n+1-1)/(n+1)

=n/(n+1).

1/n(n+1)=1/n-1/(n+1)
原式=1-1/2+1/2-1/3...-1/(n+1)
=1-1/(n+1)=n/(n+1)

1/n×(n+1)=1/n-1/(n+1)

1/1×2+1/2×3+1/3×4+…+1/n×(n+1)
=1-1/2+1/2-1/3+1/3.....+1/n-1/(n+1)
=1-1/(n+1)

1/(n(n+1))=1/n-1/(n+1)

故
1/1*2+1/2*3+1/3*4+。。。+1/(n(n+1))
=[1/1-1/2]+[1/2-1/3]+......+[1/n-1/(n+1)]
=1-1/(1+n)
=n/(n+1)