请帮忙解下这个题！在线等！1除以1×3+1除以3×5.........到2003分之1+2005分之1等于多少

1/1*3+1/3*5+........+1/2003*2005
=(1/2)*(1-1/3)+(1/2)*(1/3-1/5)+……+(1/2)*(1/2003-1/2005)
=(1/2)*(1-1/3+1/3-1/5+1/5-1/7+……+1/2003-1/2005)
=(1/2)*(1-1/2005)
=(1/2)*(2004/2005)
=1002/2005

解这种题目一般采用裂项法
先裂项，使各项可以抵消或合并
如1/3*5=1/2*(1/3-1/5)
=>
1/n(n+2)=1/2*[1/n-1/(n+2)]
1除以1×3+1除以3×5.........到2003分之1×2005分之1
=1/2*[(1-1/3)+(1/3-1/5)+……+(1/2003-1/2005)]
=1/2*(1-1/2005)
=1/2*(2004/2005)
=1002/2005

=[(1-1/3)+(1/3-1/5)+……+(1/2003-1/2005)]/2
=(1-1/2005)/2
=(2004/2005)/2
=1002/2005

1/n(n+2)=[1/n-1/(n+2)]/2

1除以1×3+1除以3×5.........到2003分之1+2005分之1
=(1/1-1/3)*1/2+(1/3-1/5)*1/2+....+(1/2003-1/2005)*1/2
=1/2(1/1-1/3+1/3-1/5+1/5-1/7+.....+1/2003-1/2005)
=1/2(1-1/2005)
=1002/2005

1/(1*3) + 1/(3*5) + ... + 1/(2001*2003) + 1/(2003*2005)
=1/2 * (1 - 1/3 + 1/3 - 1/5 + ...+ 1/2001 - 1/2003 + 1/2003 - 1/2005)
=1/2 * (1 - 1/2005)
=1002/2005<