Question

giá trị của tích (1-1/2^2)(1-1/3^2)(1-1/4^2).....(1-1/99^)(1-1/100^2)

Answer 1

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{99^2}\right)\left(1-\dfrac{1}{100^2}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{98}{99}.\dfrac{100}{99}.\dfrac{99}{100}.\dfrac{101}{100}\)

\(=\dfrac{1.2...98.99}{2.3...99.100}.\dfrac{3.4...100.101}{2.3...99.100}\)

\(=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)