Question

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

So sánh A với \(\frac{-1}{2}\)

Answer 1

ta có: \(M=-\frac{3}{2^2}.-\frac{8}{3^2}.-\frac{15}{4}...-\frac{9999}{100^2}\)

M có 99 thừa số âm

=>M<0

\(-M=\frac{3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}=>-M=\frac{\left(2.3.4...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}=\frac{101}{100.2}=\frac{101}{200}\)

\(\frac{101}{200}>\frac{100}{200}=\frac{1}{2}=>-M>\frac{1}{2}=>M