怎么简便计算
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*......*(1-100)*(1+1/100)
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*......*(1-1/100)*(1+1/100)
=1/2*3/2*2/3*4/3*......*99/100*101/100
=1/2*(3/2*2/3)*(4/3*3/4)*......*(100/99*99/100)*101/100
=1/2*1*1*1*......*1*1*101/100
=101/200
把奇数项和偶数项分别写在一起，你就发现约分的规律了。
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*......*(1-100)*(1+1/100)
=1/2 * 3/2 * 2/3 * 4/3 * …… * 99/100 * 101/100 第二项开始两两抵消，只剩首尾两项：
=1/2 * 101/100
=101/200
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*......*(1-100)*(1+1/100)
=1/2 * 3/2 * 2/3 * 4/3 * …… * 99/100 * 101/100 去掉抵消项：
=1/2 * 101/100
=101/200
回答完毕保证正确
(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)*......*(1-100)*(1+1/100)
=1/2*3/2*2/3*4/3*3/4*.........*99/100*101/100
=101/200
(1-1/2)×(1+1/2)×(1-1/3)×(1+1/3)×......×(1-100)×(1+1/100)
=1/2×（3/2×2/3)×(4/3×3/4)×……×(100/99×99/100)×101/100
=1/2×1×1×……×1×101/100
=101/200
解：原式=（1-1/2）*（1+1/2）*（1-1/3）*（1+1/3）*.（1-1/100）*(1+1/100）
=（1/2）×(3/2)×(2/3)×(4/3)×(3/4)×(5/4)×...×(99/100)×(101/100)
=(1/2)×(101/100)
=101/200
原式
=[(1/2)(2/3)(3/4)*……*(99/100)][(3/2)(4/3)(5/4)*……*(101/100)]
=(1/100)(101/2)
=101/200
（1-1/2）*（1+1/2）*（1-1/3）*（1+1/3）*.（1-1/100）*(1+1/100）
=（1/2）×(3/2)×(2/3)×(4/3)×(3/4)×(5/4)×...×(99/100)×(101/100)
=1/2×101/100
=101/200
=0.005