Question

Tính

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{1998^2}\right)\)

Answer 1

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{1998^2}\right)\)

\(=\dfrac{2^2-1}{2.2}.\dfrac{3^2-1}{3.3}.\dfrac{4^2-1}{4.4}...\dfrac{1998^2-1}{1998.1998}\)

\(=\dfrac{\left(2-1\right)\left(2+1\right)\left(3-1\right)\left(3+1\right)\left(4-1\right)\left(4+1\right)...\left(1998-1\right)\left(1998+1\right)}{\left(2.3.4...1998\right).\left(2.3.4...1998\right)}\)

\(=\dfrac{1.2.3...1997}{2.3.4...1998}.\dfrac{3.4.5...1998}{2.3.4...1998}\)

\(=\dfrac{1}{1998}.\dfrac{1}{2}\)

\(=\dfrac{1}{3996}\)