Question

(1 -1/4)(1-1/9)(1-1/16)...(1-1/n²)

Answer 1

\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)

Do đó : \(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{n^2}\right)\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot\left(n-1\right)}{2\cdot3\cdot4\cdot...\cdot n}\cdot\frac{3\cdot4\cdot5\cdot...\cdot\left(n+1\right)}{2\cdot3\cdot4\cdot...\cdot n}\)

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