[tex]\rm Termenii~sunt ~de~forma~6k+1. \\ \\ Notez~x=6n+1 .\\ \\ 1+7+13+...+6n+1= \\ =1+6 \cdot 1+1+6 \cdot 2+1+...+6n+1= \\ =6(1+2+3+...+n)+n+1= \\ = 6 \cdot \frac{n(n+1)}{2}+n+1= \\ =3n(n+1)+(n+1)= \\ =(n+1)(3n+1). [/tex]
[tex]\rm Avem:~(n+1)(3n+1)=280,~cu~solutia~naturala~n=9. \\ \\ n=9 \Rightarrow x=6 \cdot 9 +1=55. \\ \\ \underline{Solutie}:x=55.[/tex]
