Question

请教一个数学问题！

计算：（1-1/2^2)(1-1/3^2)(1-1/4^2)···（1-1/100^2)
帮帮忙，好象要用到乘法公式．

Answer 1

(1-1/2^2) = (1+1/2)*(1-1/2) = 3/2 *1/2
(1-1/3^2) = (1+1/3)*(1-1/3) = 4/3 *2/3
(1-1/4^2) = (1+1/4)*(1-1/4) = 5/4 *3/4
...
(1-1/100^2) = (1+1/100)*(1-1/100) =101/100 *99/100

所以，
（1-1/2^2)(1-1/3^2)(1-1/4^2)···（1-1/100^2)
= (3/2 *1/2)*(4/3 *2/3)*(5/4 *3/4)*...*(101/100 *99/100)
= 1/2 *101/100
= 101/200

Answer 2

你那是什么运算符号^

Answer 3

(1-1/2×2)×(1-1/3×3)×(1-1/4×4)×(1-1/5×5)×……×(1-1/100×100)=(1-1/4)×(1-1/9)×(1-1/16)×(1-1/25)……(1-1/10000)=(1-1/2)(1+1/2)(1-1/4)(1+1/4)……(1-1/100)(1+1/100)=101/200
平方差公式:(a+b)(a-b)=a的平方-B的平方得出了上面的结果,

Answer 4

1-1/2^2) = (1+1/2)*(1-1/2) = 3/2 *1/2
(1-1/3^2) = (1+1/3)*(1-1/3) = 4/3 *2/3
(1-1/4^2) = (1+1/4)*(1-1/4) = 5/4 *3/4
...
(1-1/100^2) = (1+1/100)*(1-1/100) =101/100 *99/100

所以，
（1-1/2^2)(1-1/3^2)(1-1/4^2)···（1-1/100^2)
= (3/2 *1/2)*(4/3 *2/3)*(5/4 *3/4)*...*(101/100 *99/100)
= 1/2 *101/100
= 101/200 可能是这样的！

Answer 5

楼上几位朋友的回答都是正确的，此题的关键在于利用平方差公式化简。