谁帮我做一道数学题在线等急！

裂项相消法
因为2/n*(n+2)=1/n-1/(n+2)
所以2/1*3+2/3*5+2/5*7+......+2/2007*2009
=(1-1/3)+(1/3-1/5)+(1/5-1/7)+......+(1/2007-1/2009)
=1-1/2009
=2008/2009

2/1*3+2/3*5+2/5*7+......+2/2007*2009
=2*1/2*(1/1*3+1/3*5+1/5*7+......+1/2007*2009)
=(1/1-1/3+1/3-1/5+1/5-1/7+...+1/2007-1/2009)
=(1/1-1/2009)
=2008/2009

2/1*3+2/3*5+2/5*7+......+2/2007*2009=
(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/2007-1/2009)=1-1/2009+2008/2009