Question

chứng tỏ rằng 3^1 + 3^2 + 3^3 + ... + 3^60 chia hết cho 13

Answer 1

\(3+3^2+3^3+...+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+...+3^{58}\right)⋮13\)

Answer 2

Lời giải:

Ta có: $3^1+3^2+3^3+....+3^{60}=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{58}+3^{59}+3^{60})$

$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{58}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{58})$

$=13(3+3^4+....+3^{58})\vdots 13$