1*3*5分之一加3*5*7分之一加5*7*9分之一加.加2001*2003*2005分之一等于多少

因为：
4/[(n-2)n(n+2)]
=[(n+2)-(n-2)]/[(n-2)n(n+2)]
=(n+2)/[(n-2)n(n+2)]-(n-2)/[(n-2)n(n+2)]
=1/[(n-2)n]-1/[n(n+2)]
所以：
1/(1×3×5)=(1/4)×4/(1×3×5)=(1/4)×[1/(1×3)-1/(3×5)]
1/(3×5×7)=(1/4)×4/(3×5×7)=(1/4)×[1/(3×5)-1/(5×7)]
1/(5×7×9)=(1/4)×4/(5×7×9)=(1/4)×[1/(5×7)-1/(7×9)]
……
1/(1999×2001×2003)=(1/4)×4/(1999×2001×2003)
=(1/4)×[1/(1999×2001)-1/(2001×2003)]
1/(2001×2003×2005)=(1/4)×4/(2001×2003×2005)
=(1/4)×[1/(2001×2003)-1/(2003×2005)]
那么：原式
=(1/4)×[1/(1×3)-1/(3×5)]+(1/4)×[1/(3×5)-1/(5×7)]
+(1/4)×[1/(5×7)-1/(7×9)]+……+（1/4)×[1/(1999×2001)
-1/(2001×2003)]+(1/4)×[1/(2001×2003)-1/(2003×2005)]
=(1/4)×[1/(1×3)-1/(3×5)+1/(3×5)-1/(5×7)+1/(5×7)-1/(7×9)
+……+1/(1999×2001)-1/(2001×2003)+1/(2001×2003)-1/(2003×2005)]
=(1/4)×[1/(1×3)-1/(2003×2005)]
=1004003/12048045