Question

Tính nhanh :

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{100^2}\right)\)

Giúp mình với !

Answer 1

Nhận xét :\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n+1\right)\left(n-1\right)}{n^2}\)

Do đó : \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{100^2}\right)\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.........\frac{99.101}{100^2}\)

\(=\frac{\left(1.2.3....99\right)\left(3.4.5....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)

\(=\frac{101}{100.2}\)

\(=\frac{101}{200}\)