Question

So sánh B và \(\frac{-1}{2}\)

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

Answer 1

Đặt \(100=n\) , ta có :

\(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{n^2}-1\right)\)

    \(=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.....\frac{\left(1-n\right)\left(1+n\right)}{n^2}\)

    \(=\frac{\left(-1\right).\left(-2\right)....\left(1-n\right)}{2.3.....n}.\frac{3.4........\left(1+n\right)}{2.3.....n}\)

     \(=\frac{\left(-1\right).2.3.....\left(n-1\right)}{2.3......n}.\frac{3.4.....\left(n+1\right)}{2.3.......n}\)

     \(=\frac{\left(-1\right)}{n}.\frac{n+1}{2}=\frac{-1}{2}.\frac{n+1}{n}< \frac{-1}{2}\)

Vậy \(B< \frac{-1}{2}\)