Question

Arătați că 2^1 + 2^2 + …+ 2^120 se divide cu 15.

Answer 1

$1+2^1+2^2+2^3 = 15\\ \\ \\2^1+2^2+...+2^{120} = \\ \\ = 2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+2^{9}(1+2+2^2+2^3)+...+\\+2^{117}(1+2+2^2+2^3)\\ \\ =2\cdot 15+2^5\cdot 15+2^9\cdot 15+...+2^{113}\cdot 15+2^{117}\cdot 15 = \\ \\ = 15\cdot (2+2^5+2^9+...+2^{113}+2^{117}) \\ \\ a_n =4n-3\\117 = 4n-3 \Rightarrow 4n = 120 \Rightarrow n = \dfrac{120}{4} = 30 \in \mathbb{Z}\\ \\\Rightarrow 2^1+2^2+...+2^{120} = 15\cdot (2+2^5+2^9+...+2^{113}+2^{117})~~(A) \\ \\\Rightarrow2^1+2^2+...+2^{120} ~\vdots~15$