Răspuns:
C=3+3^2+3^3+...+3^2003+3^2004
avem 2004 termeni deci putem sa ii grupam cate 2
C=(3+3^2)+(3^3+3^4)...(3^2003+3^2004)=
=(3+3^2)+3^2(3+3^2)...3^2002(3+3^2)=
=12+3^2*12+...3^2002*12=
=12*(1+3^2+...3^2002)=
=3*4*(1+3^2+...3^2002) ⇒ divizibil cu 4