Question

Tính A = (1-\(\frac{1}{2^2}\))(1-\(\frac{1}{3^2}\))(1-\(\frac{1}{4^2}\))....(1-\(\frac{1}{n^2}\))

Answer 1

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

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

\(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}....\cdot\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)

\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n-1}{n}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\frac{n+1}{n}\right)\)

\(=\frac{1}{n}\cdot\frac{n+1}{2}=\frac{n+1}{2n}\)