计算:(1+1⼀2)*(1-1⼀2)*(1+1⼀3)*(1-1⼀3)*......*(1+1⼀100)*(1-1⼀100),说的详细点!

2024-12-23 11:05:24
推荐回答(3个)
回答1:

(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)......(1+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*3/4*5/4*....*99/100*101/100从第二项开始,相邻两项互为倒数
=1/2*101/100
=101/200

回答2:

将加的相乘 再乘以 减的相乘
(1+1/2)(1+1/3)(1+1/4)……(1+1/100)×(1-1/2)(1-1/3)(1-1/4)……(1-1/100)
=3/2*4/3*5/4……101/100×1/2*2/3*3/4*……*99/100
=101/2×1/100
=101/200

回答3:

先找规律吧
(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)=3/4 * 8/9=2/3
(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*(1+1/4)*(1-1/4)=2/3 * 15/16=5/8=2.5/4
(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*.......*(1+1/5)*(1-1/5)=5/8 * 24/25=3/5
(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*......*(1+1/6)*(1-1/6)=3/5 * 35/36=7/12=3.5/6
由以上规律可以发现,
(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*......(1+1/n)*(1-1/n)
=(n/2+0.5)/n
所以:(1+1/2)*(1-1/2)*(1+1/3)*(1-1/3)*......*(1+1/99)*(1-1/99)
=(99/2+0.5)/99 =50/99