1/1x3+1/3x5+1/5x7+...+1/2007x2009=?

1/1x3+1/3x5+1/5x7+...+1/2007x2009=?
1/1x3+1/3x5+1/5x7+...+1/2007x2009=?

1/1x3+1/3x5+1/5x7+...+1/2007x2009=?
1/1x3+1/3x5+1/5x7+...+1/2007x2009
=(1-1/3)÷2+(1/3-1/5)÷2+(1/5-1/7)÷2+……+(1/2007-1/2009)÷2
=[(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+(1/2007-1/2009)]÷2
=（1-1/3+1/3-1/5+1/5-1/7+……+1/2007-1/2009）÷2
中间互相抵消
=(1-1/2009)÷2
=2008/2009÷2
=1004/2009

1/1x3+1/3x5+1/5x7+...+1/2007x2009
=1/2*(1-1/3)+1/2*(1/3-1/5)+1/2*(1/5-1/7)+...+1/2*(1/2006-1/2007)+1/2*(1/2007-1/2009)
=1/2*(1-1/3+1/3-1/5+1/5-1/7+...+1/2006-1/2007+1/2007-1/2009)
=1/2*(1-1/2009)
=1/2*(2008/209)
=1004/2009.