1+1/1+2+1/1+2+3+……+1/1+2+……+1999

来源：百度知道 编辑：UC知道 时间：2024/07/02 11:58:52
最好有步骤

An=1/(1+2+……+n)
1+2+…+n=n*(n+1)/2
An=2/[n*(n+1)]=2*[1/n - 1/(n+1)]
所以：1+1/1+2+1/1+2+3+……+1/1+2+……+1999
＝2*(1/1-1/2+1/2-1/3+…+1/1999-1/2000)
=2*(1-1/2000)
=1999/1000

an=1/[n(n+1)/2]=2/n(n+1)=2(1/n-1/(n+1))
则原式=2（1-1/2+1/2-1/3+...+1/1998-1/1999+1/1999-1/2000）
=2*(1-1/2000)
=2*1999/2000
=1999/1000

1+2/2*3+...+2/1999*2000
=1+2*(1/2-1/3+1/3-1/4+...+1/1999-1/2000)
=1+1-1/1000
=1999/1000

=1+1/3+1/6+1/10+1/15+1/21+...+1/1999000
=2(1/2+1/6+1/12+1/20+1/30+1/42+...+1/3998000)
=2(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7...+1/1999-1/2000)
= 2(1-1/2000)
=1999/1000

2000000