Question

Se considera numarul a=1 +3+3(la puterea 2)+.....+3(la puterea 101).
a)determinati daca este numar par.
b)demonstrati ca,, a ''se divide cu 13

Answer 1

a)
$1+3+3^2+...+3^{101}$
$=(1+3)+(3^2+3^3)+...+(3^{100}+3^{101})$
$=(1+3)+3^2(1+3)+...+3^{100}(1+3)$
$=4+3^2\cdot4+...+3^{100}\cdot4$
$=4(1+3^2+...+3^{100})~\vdots~2$

b)
$=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^{99}+3^{100}+3^{101})$
$=(1+3+3^2)+3^3(1+3+3^2)+...+3^{99}(1+3+3^2)$
$=13+3^3\cdot13+...+3^{99}\cdot13$
$=13(1+3^3+...+3^{99})~\vdots~13$