Question

Arata ca (9 la puterea 1 + 9 la puterea 2 + 9 la puterea 3 +.... + 9 la puterea 2012) divizibil cu 10

Answer 1

$Fie~a=9^{1}+9^{2}+9^{3}+...+9^{2012}=\\\\(avem~2012~termeni~pe~care~ii~vom~grupa~cate~doi)\\\\ =(9^{1}+9^{2})+(9^{3}+9^{4})+...+(9^{2011}+9^{2012})=\\\\=(9^{1}+9^{1+1})+(9^{3}+9^{3+1})+...+(9^{2011}+9^{2011+1})=\\\\Dam~factor~comun~primul~termen~al~fiecarei~paranteze\\\\=9^{1}(1+9^{1})+9^{3}(1+9^{1})+...+9^{2011}(1+9^{1})=\\\\=9^{1}\cdot 10 +9^{3}\cdot 10 +9^{5}\cdot10+...+9^{2011}\cdot 10=\\\\Dam~factor~comun~pe~10\\\\=10\cdot(9^{1}+9^{3}+9^{5}+...+9^{2011})=\\\\ \Rightarrow a~divizibil~cu~10$