“(1-1/22)(1-1/32)(1-1/42)…(1-1/20052)(1-1/20062)(1-1/20072)(1-1/20082)(1-1/20092)”是否是“(1-1/2²)(1-1/3²)(1-1/4²)…(1-1/2005²)(1-1/2006²)(1-1/2007²)(1-1/2008²)(1-1/2009²)”?若是,可用下面的方法求
(1-1/2²)(1-1/3²)(1-1/4²)…(1-1/2005²)(1-1/2006²)(1-1/2007²)(1-1/2008²)(1-1/2009²)
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)(1-1/5)(1+1/5)(1-1/6)(1+1/6)(1-1/7)(1+1/7)(1-1/8)(1+1/8)……(1-1/2005)(1+1/2005)(1-1/2006)(1+1/2006)(1-1/2007)(1+1/2007)(1-1/2008)(1+1/2008)(1-1/2009)(1+1/2009)
=(1-1/2)(1+1/2009)
=1005/2009.