Question

Radical din 1 +radical din 1+3+ radical din 1+3+5 + radical din 1+3+5+...+(2x+1)= 201x10055

Answer 1

[tex]1+3+5.+..+(2n-1)=n^2~este~o~formula~cunoscuta\\ \sqrt1=\sqrt{1^2}=\bold{1}\\ \sqrt{1+3}=\sqrt4=\bold2\\ \sqrt{1+3+5}=\sqrt9=\bold3\\ ....\\ \sqrt{1+3+5+...+(2x+1)}=\sqrt{1+3+5+...+(2(x+1)-1)}=\\ =\sqrt{(x+1)^2}=\bold{x+1} \\ Ecuatia~devine:\\ 1+2+3+...+(x+1)=201\cdot10055\\ \frac{(x+1)(x+2)}{2}=201\cdot10055\\ Aceasta~ecuatie~de~gradul~al~doilea~are ~solutia~naturala~\boxed{x=2009}[/tex]