1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+.....+2004)=?

来源：百度知道 编辑：UC知道 时间：2024/05/22 13:38:02

问题补充：1+1/(1＋2)＋1/(1＋2+3)……+1/(1＋2+3……+2004)

1/(1+2+3+....n)=1/[(1+n)n/2]=2/[n(n+1)]=2[1/n-1/(n+1)]

所以原式=1+2(1/2-1/3)+2(1/3-1/4)+...+2(1/2004-1/2005)

=1+2(1/2-1/3+1/3-1/4+...+1/2004-1/2005)

=1+2(1/2-1/2005)

=1+1-2/2005

=4008/2005

=1+1-1/2+1/2-1/3+.......-1/2004=4007/2004

An=1/(1+2+3+...n)=2/(n(n+1))
1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+.....+2004)
=2/(1*2)+2/(2*3)+2/(3*4)+.....2/(2004*2005)
=2*(1/1-1/2+1/2-1/3+......1/2004-1/2005)
=2*(1-1/2005)
=4008/2005

1+2+……+n=n(n+1)/2
所以1/(1+2+……+n)=2/n(n+1)=2*[1/n-1/(n+1)]

所以1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+.....+2004）
=2*{(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1*2004-1/2005)]
=2*(1/1-1/2005)
=4008/2005

出这种题的都属于神经系的