1、(1/2005 - 1)(1/2004 -1)……(1/3 -1)(1/2-1)
1、(1/2005 - 1)(1/2004 -1)……(1/3 -1)(1/2-1)
原式=(-1)^2004*(1-1/2005)(1-1/2004)*......*(1-1/2)
=(1-1/2005)(1-1/2004)*......*(1-1/2)
=(2004/2005)*(2003/2004)*.....*(1/2)
=(2004*2003*...*1)/(2005*2004*....*2)
=1/2005
一共是偶数个项相乘, 结果等同于
1-1/2)(1-1/3)...(1/1/2004)(1-1/2005) =(1/2)*(2/3)*(3/4)*...*(2003/2004)*(2004/2005) =1/2005
(1/2005 - 1)(1/2004 -1)……(1/3 -1)(1/2-1)
=(-2004/2005)(-2003/2004).........(-2/3)*(-1/2)
=1/2005
解:提个-1得原式=2004/2005×2003/2004×2002/2003×……2/3×1/2
=1/2005