Question

cứu đi,gấp lắm

 

Answer 1

Ta có: \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\cdot\ldots\cdot\left(\frac{1}{100^2}-1\right)\)

\(=\left(\frac12-1\right)\left(\frac13-1\right)\cdot\ldots\cdot\left(\frac{1}{100}-1\right)\left(\frac12+1\right)\left(\frac13+1\right)\cdot\ldots\cdot\left(\frac{1}{100}+1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\ldots\cdot\frac{-99}{100}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{101}{100}\)

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