Question

Bài 1. Tính S=\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{100^2}\right)\)

Answer 1

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

\(=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}\)

\(=\frac{\left(1.3\right).\left(2.4\right).........\left(99.101\right)}{\left(2.2\right).\left(3.3\right).......\left(100.100\right)}\)

\(=\frac{\left(1.2....99\right)\left(3.4....101\right)}{\left(2.3.....100\right)\left(2.3....100\right)}\)

\(=\frac{1.101}{100.2}=\frac{101}{200}\)