Question

TÍnh: \(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)

Answer 1

ta có: n² - 1 = (n² - n) + (n -1) = n(n-1) + (n-1) = (n-1).(n+1) ; n \(\in\) N

  Áp dụng công thức tổng quát trên ta có: 

A = (\(\dfrac{1}{2^2}\) - 1).(\(\dfrac{1}{3^2}\) - 1)...(\(\dfrac{1}{100^2}\) - 1)

A = \(\dfrac{2^2-1}{-2^2}\). \(\dfrac{3^2-1}{-3^2}\)......\(\dfrac{100^2-1}{-100^2}\)

A =  \(\dfrac{\left(2-1\right)\left(2+1\right)}{-2^2}\).\(\dfrac{\left(3-1\right).\left(3+1\right)}{-3^2}\).....\(\dfrac{\left(100-1\right).\left(100+1\right)}{-100^2}\)

A = - \(\dfrac{1.3.2.4.3.5.......99.101}{2^2.3^2.4^2...100^2}\)

A = - \(\dfrac{101}{200}\)