化简代数式

用平方差
原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)……(1-1/2006)(1+1/2006)
=(1/2)(3/2)(2/3)(4/3)……(2005/2006)(2007/2006)
中间约分约去
=(1/2)(2007/2006)
=2007/4012

1-1/{(n+1)^2}==n(n+2)/{(n+1)^2}
原式==
(1*3)/(2^2)*(2*4)/(3^2)*(3*5)/(4^2).............
*（2003*2005)/(2004^2)*(2004*2006)/(2005^2)
*(2005*2007)/(2006^2)=2007/(2*2006)
==2007/4012

(1-1/2的平方)(1-1/3的平方)(1-1/4的平方).......(1-1/2006的平方)
=（1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)……(1-1/2006)(1+1/2006)
=(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)……（2005/2006)(2007/2006)
=(1/2)(2007/2006)=2007/4012