Question

Se considera nr A=$3^{0}+ 3^{1} + 3^{2}+...+ 3^{2007}$ . Aratati ca A este divizibil cu 10.

Answer 1

$A=(3^0+3^1+3^2+3^3)+(3^4+3^5+3^6+3^7)+...+(3^{2004}+ \\ +3^{2005}+3^{2006}+3^{2007})= \\ \\ ~~~=(3^0+3^1+3^2+3^3)+3^4(3^0+3^1+3^2+3^3)+...+3^{2004}(3^0+3^1+ \\ +3^2+3^3)= \\ \\ ~~~=(3^0+3^1+3^2+3^3)(1+3^4+3^7+...+3^{2004})= \\ \\ ~~~=(1+3+9+27)(1+3^4+3^7+...+3^{2004})= \\ \\ ~~~=40(1+3^4+3^7+...+3^{2004})= \\ \\ ~~~= \boxed{10} \cdot 4 \cdot (1+3^4+3^7+...+3^{2004}) \Rightarrow A ~ \vdots ~ 10.$