Question

[Nguyễn Thùy Chi]

Cho \(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)

So sánh \(A\) với \(\dfrac{-1}{2}\)

Answer 1

[Phan Công Bằng]

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)

\(=\left(\dfrac{-3}{4}\right)\left(\dfrac{-8}{9}\right)\left(\dfrac{-15}{16}\right)...\left(\dfrac{-399}{400}\right)\)

\(=\dfrac{-3.8.15...399}{4.9.16...400}\)

\(=\dfrac{-3.2.4.3.5...21.19}{2^2.3^2.4^2...20^2}\)

\(=\dfrac{-2.3.4...19}{2.3.4...20}.\dfrac{3.4.5...21}{2.3.4...20}\)

\(=\dfrac{-1}{20}.\dfrac{21}{2}\)

\(=\dfrac{-21}{40}< \dfrac{-1}{2}\)

Vậy \(A< \dfrac{-1}{2}\)