Răspuns :
Grupam termenii sumei 2 cate 2 si dam factor comun
[tex]a=(1+5)+5^{2}(1+5)+5^{4}(1+5)+..+5^{2010}(1+5)+5^{2012}(1+5)=6+5^{2}*6+5^{4}*6+...+5^{2010}*6+5^{2012}*6=6(1+5^{2}+5^{4}+5^{6}+...+5^{2010}+5^{2012})[/tex] care este evident divizibil cu 6
[tex]a=(1+5)+5^{2}(1+5)+5^{4}(1+5)+..+5^{2010}(1+5)+5^{2012}(1+5)=6+5^{2}*6+5^{4}*6+...+5^{2010}*6+5^{2012}*6=6(1+5^{2}+5^{4}+5^{6}+...+5^{2010}+5^{2012})[/tex] care este evident divizibil cu 6
[tex]a= 1+5+5^{2}+ 5 ^{3} ......+5^{2013} \\ \\ a=(1+5)+5 ^{2} (1+5)+....5 ^{2012} (1+5) \\ \\ a=6(1+5 ^{2} +...+5 ^{2012} ):6 [/tex]
este divizibil cu 6
este divizibil cu 6