(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)...*(1-1/99)*(1+1/99)
来源:百度知道 编辑:UC知道 时间:2024/06/23 11:59:15
是高手就来!有过程哦~~
答案是:50/99
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)...*(1-1/99)*(1+1/99)
先算各个括号里的
=(1/2)*(3/2)*(2/3)*(4/3)*(3/4)*(5/4)*.....(98/99)*(100/99)
=分子与分母相同的就相抵消
最后剩下
=(1/2)*(100/99)
=50/99
这种题,首先你把前几步写出来,找规律
方法就那几种,看多了,做多了就熟练了
原式=[(1-1/2)*(1-1/3)*...*(1-1/99)]
*[(1+1/2)*(1+1/3)...*(1+1/99)]
=(1/2*2/3*...*98/99)*(3/2*4/3*...*100/99)=(1/99)*(100/2)
=50/99.
支持二楼朋友的做法。
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1+1/1+2+1/1+2+3.........+1/1+2+3.....100
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
1+1/2+1+1/3+1+1/4+......+1/100=?