问两道数学题，急！

来源：百度知道 编辑：UC知道 时间：2024/09/22 05:20:18
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+...+2009)=

(1/2-1/4+1/6-1/8+/1/10-1/12+...+1/198-1/200)/(1/51+1/52+1/53+1/54+…+1/100)=
满意再悬赏！！！

1+2+3+4+...+n=n(n+1)/2,
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+...+n)
=1+2/(2*3)+2/(3*4)+...+2/(2009*2010)
=1+2[(1/2-1/3)+(1/3-1/4)+...(1/2009-1/2010)]
=1+2*(1/2-1/2010)
=1+1004/1005
=2009/1005
第二题很抱歉~~

设an=1/(1+2+3+...+n)=1/[n(n+1)/2]=2/[n(n+1)]=2/n-2/(n+1),
则原式=a1+a2+a3+...+a2008+a2009
=2/1-2/2+2/2-2/3+2/3-2/4+...+2/2008-2/2009+2/2009-2/2010
=2/1-2/2010
=2009/1005