D=1 +3+3^2+3^3+...+3^2021+3^2022 CM D -1 chia hết cho 13

Giải thích các bước giải:

$\text{Ta có: D = 1 + 3 + 3² + 3³ +...+ $3^{2021}$ + $3^{2022}$}$

$\text{⇒ D - 1 = 3 + 3² + 3³ +...+ $3^{2021}$ + $3^{2022}$}$

$\text{⇒ D - 1 = 3 + 3² + 3³ +...+ $3^{2021}$ + $3^{2022}$}$

$\text{⇒ D - 1 = (3 + 3² + 3³) +...+ ($3^{2020}$ + $3^{2021}$ + $3^{2022}$)}$

$\text{⇒ D - 1 = 3(1 + 3 + 3²) +...+ $3^{2020}$(1 + 3 + 3²)}$

$\text{⇒ D - 1 = 3 . 13 +...+ $3^{2020}$ . 13}$

$\text{⇒ D - 1 = 13(3 +...+ $3^{2020}$)}$

$\text{Vì 13 $\vdots$ 13 ⇒ 13(3 +...+ $3^{2020}$) $\vdots$ 13}$

$\text{ Hay D - 1 $\vdots$ 13}$

$\textit{@Ngốc}$