Ta có:
\(\left(1-\dfrac{1}{2^2}\right).\left(1-\dfrac{1}{3^2}\right).\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{n^2-1}{n^2}\)
\(=\dfrac{3.8.15....\left(n^2-1\right)}{4.9.16.....n^2}\)
\(=\dfrac{1.3.2.4.3.5....\left(n-1\right)\left(n+1\right)}{2.2.3.3.4.4....n.n}\)
\(=\dfrac{\left[1.2.3....\left(n-1\right)\right].\left[3.4.5....\left(n+1\right)\right]}{\left(2.3.4....n\right).\left(2.3.4....n\right)}\)
\(=\dfrac{1.\left(n+1\right)}{n.2}=\dfrac{n+1}{2n}\)
Ta có công thức:
\(1-\dfrac{1}{k^2}=\dfrac{k^2-1^2}{k^2}=\dfrac{\left(k+1\right)\left(k+2\right)}{k^2}\)
Áp dụng công thức trên ta đc:
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{2^2-1^2}{2^2}.\dfrac{3^2-1^2}{3^2}.\dfrac{4^2-1^2}{4^2}....\dfrac{n^2-1^2}{n^2}\)
\(=\dfrac{\left(2+1\right)\left(2-1\right)}{2.2}.\dfrac{\left(3+1\right)\left(3-1\right)}{3.3}.\dfrac{\left(4+1\right)\left(4-1\right)}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{[1.2.3....\left(n+1\right)].[3.4.5....\left(n-1\right)]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)
\(=\left(n+1\right).\dfrac{1}{2n}=\dfrac{n+1}{2n}\)
Chúc bạn học tốt!
