Question

[Ngân Nguyễn]

Tính :

D=\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)..........\left(1-\frac{1}{n^2}\right)\)

Answer 1

[Stephen Hawking]

\(D=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)......\left(1-\frac{1}{n^2}\right)\)

    \(=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}........\frac{n^2-1}{n^2}\)

    \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.......\frac{\left(n-1\right)\left(n+1\right)}{n.n}\)

   \(=\frac{1.3.2.4.3.5......\left(n-1\right)\left(n+1\right)}{2.2.3.3.4.4......n.n}\)

   \(=\frac{[1.2.3......\left(n-1\right)].[3.4.5......\left(n+1\right)]}{\left(2.3.4......n\right)\left(2.3.4......n\right)}\)

   \(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)

Vậy \(D=\frac{n+1}{2n}\)