Question

3 la puterea 1 + 3 la puterea 2 +........3 la puterea 99 este divizibil cu 5

Answer 1

$3^1+3^2+3^3+3^4+...+ 3^{96}+3^{97}+3^{98}+3^{99}= \\\\ =(3+9+27+81)+....+3^{95}(3+3^2+3^3+3^4)= \\\\ = 120+...+3^{95}*120= \\\\ = 120(1+....+3^{95})= \\\\=5* 24(1+...+3^{95}) \ \ \vdots \ \ 5 \ \ \ \ \hbox{(c.c.t.d)}$