Question

1+5+$5^{2} + 5^{3} + 5^{4} +...+ 5^{101} divizibil cu 31.$

Answer 1

cu A notez suma:
$A=5^{0}+5^{1}+5^{2}+5^{3}+5^{4}+......+5^{101}$
numarul de termeni de la 0 la 101=102 termeni=numar par de termeni
Grupam termenii cate 3 si obtinem 102/3=34 grupe
$A=(5^{0}+5^{1}+5^{2})+(5^{3}+5^{4}+5^{5})+......+(5^{99}+5^{100}+5^{101})$
$A=(5^{0}+5^{1}+5^{2})*(5^{0})+(5^{0}+5^{1}+5^{2})*(5^{3})+...+(5^{0}+5^{1}+5^{2})*(5^{99})$
$A=(5^{0}+5^{1}+5^{2})*(5^{0}+5^{3}+...+5^{99}) \\ A=31*(5^{0}+5^{3}+...+5^{99})$
Deducem ca A este divizibil cu 31.