Question

Tính :   ( \(\frac{1}{2^2}\) - 1 ) .( \(\frac{1}{3^2}\)-1 ) .( \(\frac{1}{4^2}\)-1 ) ........ ( \(\frac{1}{98^2}\)-1 ) .( \(\frac{1}{99^2}\)-1 )

Answer 1

A là tích của 98 số âm. Do đó :

\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{99^2}\right)=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{9800}{99^2}=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{98.100}{99^2}=\frac{1.2.3...97.98}{2.3.4...98.99}.\frac{3.4.5...99.100}{2.3.4...98.99}=\frac{1}{100}\cdot\frac{100}{2}=\frac{1}{2}\)

Vậy A = \(\frac{-1}{2}\)