Ta có :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(\Leftrightarrow A=-\frac{3}{4}.-\frac{8}{9}....-\frac{9999}{10000}\)
\(\Leftrightarrow A=\frac{-3.8....9999}{4.9.10000}=\frac{-3.2.4.....99.101}{2.2.3.3....100.100}=\frac{-101}{100}\)
Mà \(-\frac{1}{2}=-\frac{50}{100}>-\frac{101}{100}\)
Vậy A < \(-\frac{1}{2}\)
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