Question

b Arătaţi că B= 2° +2° +2° +2° +2° +2° +2° + ... + 2003 se divide cu 15.

Answer 1

Răspuns: Demonstrația mai jos

Explicație pas cu pas:

$\bf B=2^0 + 2^1 +2^2 +2^{3}+2^{4}+2^{5} + ....+2^{2003}$

$\bf B=\Big(2^0 + 2^1 +2^2 +2^{3}\Big)+\Big(2^{4}+...+2^{7}\Big)+ ...+\Big(2^{2000}+... +2^{2003}\Big)$

$B=\Big(1+ 2 +4 +8\Big)+2^{4}\cdot\Big(2^{4-4}+...+2^{7-4}\Big)+ ...+2^{2000}\cdot\Big(2^{2000-2000}+... +2^{2003-2000}\Big)$

$\bf B=15+2^{4}\cdot\Big(2^{0}+2^1 +2^2 +2^{3}\Big)+ ...+2^{2000}\cdot\Big(2^{0}+2^1 +2^2 +2^{3}\Big)$

$\bf B=15+2^{4}\cdot\Big(1+ 2 +4 +8\Big)+ ...+2^{2000}\cdot\Big(1+ 2 +4 +8\Big)$

$\bf B=15+2^{4}\cdot 15+ ...+2^{2000}\cdot 15$

$\pink{\boxed{~\bf B=15\cdot\Big(2^{0}+2^{4}+2^{8}+... +2^{2000}\Big)~~\vdots~~15~}}$

$==pav38==$