简单的数学题

1 1 1 1

--- + --- + --- … ---------

1*2 2*3 3*4 2005*2006

怎么做啊

用裂项相消法

根据 1/n*(n+1)= 1/n - 1/(n+1)

原式=1- (1/2)+(1/2)-(1/3)+(1/3)-(1/4)+...+(1/2005)-(1/2006)

=1-(1/2006)

=2005/2006