Question

Arata ca numarul S=1 supra 2+1 supra 6+...+1 supra 9900 este subunitar.

Answer 1

[tex]S=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\\ \\ S=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+....+\frac{1}{99\cdot 100}\\ \\ S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\\ \\ S=1-\frac{1}{100}\\ S=\frac{99}{100}\\ Cum\ 99\ \textless \ 100\Rightarrow Fractia\ este\ subunitara.[/tex]