Giải thích các bước giải:
Ta có: \(\frac{1}{n^2}-1 = \frac{1-n^2}{n^2}= \frac{(1-n)(1+n)}{n^2}\)
Suy ra:
A = \(\frac{-1.3}{2^2}. \frac{-2 . 4}{3^2}.\frac{-3.5}{4^2}....\frac{-99.101}{100^2}\)
\(= -\frac{(1 \times 2 \times ....\times 99)( 3 \times 4 \times 5...\times 101)}{( 2\times 3\times 4...\times 100)^2}\)
= \(- \frac{99!.\frac{101!}{2}}{(100!)^2}\)
= \(- \frac{101}{100.2}<\frac{-100}{100.2}=\frac{-1}{2}\)
Nên \(A<\frac{-1}{2}\)
