(1-1/2^2)(1-1/3^2)(1-1/4^2).....

[iCantPlayPudge]

(1-1/2^2)(1-1/3^2)(1-1/4^2)(1-1/5^2)(1-1/6^2)(1-1/7^2)(1-1/8^2)(1-1/9^2)(1-1/10^2)

Da se presmetne bez kalkulator i bez operacii

[Добромир Глухаров]

[tex]1-\frac{1}{k^2}=\(1-\frac{1}{k}\)\(1+\frac{1}{k}\),\ k=2,3,...,10[/tex]

Получава се:

[tex]\(1-\frac{1}{2}\)\(1+\frac{1}{2}\)\(1-\frac{1}{3}\)\(1+\frac{1}{3}\)\(1-\frac{1}{4}\)\(1+\frac{1}{4}\)\cdots\(1-\frac{1}{9}\)\(1+\frac{1}{9}\)\(1-\frac{1}{10}\)\(1+\frac{1}{10}\)=[/tex]

[tex]=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdots\frac{8}{9}\cdot\frac{10}{9}\cdot\frac{9}{10}\cdot\frac{11}{10}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}[/tex]