Question

Aratati ca numarul urmator este rational
B=n(n+1)(n+2)(n+3)+1
Totul sub radical

Answer 1

[tex] \sqrt[]{n(n+1)(n+2)(n+3)+1}= \sqrt[]{n(n+3)(n+1)(n+2)+1}=\\ =\sqrt{(n^2+3n)(n^2+3n+2)+1}=\\ \sqrt{(n^2+3n+1-1)(n^2+3n+1+1)+1}=\\ \sqrt{((n^2+3n+1)-1)((n^2+3n+1)+1)+1}=\\ =\sqrt{(n^2+3n+1)^2-1^1+1}=\sqrt{(n^2+3n+1)^2}=\\ =(n^2+3n+1)\in\mathbb{Q}.\\[/tex]