急!!数学题，在线等。

裂项求和1/[n*(n+2)]=1/2*[1/n-1/(n+2)]
原式=1/2*(1/1-1/3+1/3-1/5+1/5-1/7+……+1/2007-1/2009)
=1/2*(1-1/2009)=1004/2009

原式=1/2*（1/1-1/3+1/3+1/5-1/5+1/7+……+1/2007-1/2009）=1/2*（1-1/2009）=1004/2009=下边的会算了吧？

1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+....1/2007*2009=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+....+1/2007-1/2009)=1/2(1-1/2009)=1004/2009

1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7）……+1/2(1/2007-1/2009)=1/2(1-1/3+1/3-1/5+1/5-1/7……+1/2007-1/2009)=1/2(1-1/2009)=1004/2009

原式=1/2(1-1/3 +1/3-1/5 +1/5-1/7+......+1/2007-1/2009)
=1/2(1-1/2009)
=1004/2009

1/2*(1/1-1/3+1/3-1/5+1/5-1/7+.....+1/2007-1/2009)=1/2*(1-1/2009)=1004/2009