Question

So sánh A với \(-\dfrac{1}{2}\) biết:

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

Answer 1

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

\(=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)\) ( do có 99 cặp số )

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

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

\(=-\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)

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

Vậy \(A< \dfrac{-1}{2}\)

Answer 2

Đề sai rồi kìa bn

Đúg ra phải là 1/100^2 -1 chứ