x=2 y=4,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+...+1/(x+2010)(y+2010)的值怎么算出
问题描述：
x=2 y=4,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+...+1/(x+2010)(y+2010)的值怎么算出
答
原式=1/(2*4) + 1/(4*6) +……+1/(2012*2014)
= (1/2 - 1/4)/2 + (1/4 -1/6)/2 +……+(1/2012 - 1/2014)/2
= (1/2 -1/2014) /2
=1003/2014
答
x=2 y=4,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+...+1/(x+2010)(y+2010)原式= 1/(2*4) + 1/(4*6) + ... + 1/(2012*2014)= (1/2)(1/2-1/4+1/4-1/6+...+1/2012-1/2014)= (1/2)(1/2 - 1/2014) (中间全都约掉了)= 503/2014...