Question

[Lê Quỳnh Trang]

\(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\cdot\cdot\left(\frac{1}{100^2}-1\right)\)

So sánh A với \(-\frac{1}{2}\)

Answer 1

[Không Tên]

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

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

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

     \(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

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

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

=>  \(A=-\frac{101}{200}< -\frac{1}{2}\)