计算:1*3分之一减3*5分之一……减2005*2007分之一

因为1/[n(n+2)]=(1/2)×[1/n-1/(n+2)]
1/(1×3)+1/(3×5)+1/(5×7)+...+1/(2003×2005)+1/(2005×2007)
=(1/2)×(1/1-1/3)+(1/2)×(1/3-1/5)+(1/2)×(1/5-1/7)+...+(1/2)×(1/2003-1/2005)+(1/2)×(1/2005-1/2007)
=(1/2)×(1-1/3+1/3-1/5+1/5-1/5+...+1/2003-1/2005+1/2005-1/2007)
=(1/2)×(1-1/2007)
=(1/2)×2006/2007
=1003/2007
希望能够帮助你,有疑问欢迎追问,祝学习进步!连接符号是减不是加因为1/[n(n+2)]=(1/2)×[1/n-1/(n+2)]1/(1×3)-1/(3×5)-1/(5×7)-...-1/(2003×2005)-1/(2005×2007)=(1/2)×(1/1-1/3)-(1/2)×(1/3-1/5)-(1/2)×(1/5-1/7)-...-(1/2)×(1/2003-1/2005)-(1/2)×(1/2005-1/2007)=(1/2)×(1-1/3-1/3+1/5-1/5+1/7-...-1/2003+1/2005-1/2005+1/2007)=(1/2)×(1-1/3-1/3+1/2007)=(1/2)×(1/3+1/2007)=(1/2)×670/2007=335/2007不好意思，我为我的疏忽道歉，祝学习进步！