Question

Tính : \(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)

Answer 1

Ta có: \(1-\dfrac{1}{k^2}=\dfrac{k^2-1}{k^2}=\dfrac{\left(k+1\right)\left(k+1\right)}{k^2}\) nên:

\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{99.101}{100^2}=\dfrac{1.3.2.4.....99.101}{2^2.3^2.4^2....100^2}=\dfrac{1.2.3.....99}{2.3.4.5....99.100}.\dfrac{3.4.5.....101}{2.3.4.5.....100}=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)

Chúc bạn học tốt !!!

Answer 2

WTF?

Answer 3

Ta có

A = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)(có 99 thừa số)

A = 1.1.1.1.1.1...1 - (\(\dfrac{1}{2^2}.\dfrac{1}{3^2}.\dfrac{1}{4^2}....\dfrac{1}{100^2}\))

A = 1 - \(\dfrac{1.1.1.1.1....1}{2^2.3^2.4^2...100^2}\)

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

Vậy A = \(1-\dfrac{1}{100!^2}\)