Question

(1+1/2)乘（1-1/2）乘（1+1/3）乘（1-1/3）乘...乘（1+1/99）乘（1-1/99）= 求详解

Answer 1

利用交换律，将“-”的为一组，“+”的为一组
(1+1/2)×(1-1/2)×(1+1/3)×(1-1/3)……(1+1/99)×(1-1/99)
= [(1+1/2)×(1+1/3)×(1+1/4)×……×(1+1/99)]×[(1-1/2)×(1-1/3)×(1-1/4)×……×(1-1/99)]
=[(3/2)×(4/3)×(5/4)×……×(100/99)]×[(1/2)×(2/3)×(3/4)……(98/99)]
=(100/2)×(1/99)=50/99

Answer 2

解：因为（1+1/n)(1-1/n)=(n+1)(n-1)/n^2（n>=2)
所以(1+1/2)乘（1-1/2）乘（1+1/3）乘（1-1/3）乘...乘（1+1/99）乘（1-1/99）= (3*1/2^2)*(4*2/3^2)*……*（100*98/99^2）=50/98

Answer 3

=(1-1/2)(1+1/2)(1-1/3)(1+1/3)……(1-1/99)(1+1/99)
=(1/2)(3/2)(2/3)(4/3……(98/99)(100/99)
中间约分
=(1/2)(100/99)
=50/99