(2*4)/1+(4*6)/1(6*8)/1.....+(198*200)/1=

来源：百度知道 编辑：UC知道 时间：2024/06/08 14:50:38
最好是能讲一讲做这道题的具体方法，拜托各位能人了！

1/(2*4)+1/(4*6)+1/(6*8)+...+1/(198*200)
=(1/2)*[(1/2)-(1/4)]+(1/2)[(1/4)-(1/6)]+(1/2)[(1/6)-(1/8)]+...+(1/2)[(1/198)-(1/200)]
=(1/2)[(1/2)-(1/4)+(1/4)-(1/6)+...+(1/198)-(1/200)]
=(1/2)[(1/2)-(1/200)]
=(1/2)*(199/200)
=199/400.

题目应该是：1/(2*4)+1/(4*6)+1/(6*8).....+1/(198*200)=

然后解答如下：
原式=1/4*[1/(1*2)+1/(2*3)+1/(3*4).....+1/(99*100)]
=1/4*[1-1/2+1/2-1/3+1/3-1/4.....+1/99-1/100]
=1/4*[1-1/100]
=99/400

是不是1/(2*4)+1/(4*6)+1/(6*8)+……+1/(198*200)??
如果是，则
原式=[2/(2*4)+2/(4*6)+……+2/(198*200)]/2
=[(1/2-1/4)+(1/4-1/6)+……+(1/198-1/200)]/2
=(1/2-1/200)/2
=99/400