Đáp án: $3$
Giải thích các bước giải:
Ta có:
$A=1+3+3^2+3^3+...+3^{2021}$
$\to A=1+3+(3^2+3^3)+...+(3^{2020}+3^{2021})$
$\to A=4+3(3+3^2)+...+3^{2019}(3+3^{2})$
$\to A=4+(3+3^2)(3+...+3^{2019})$
$\to A=4+12(3+...+3^{2019})$
$\to A\quad\vdots\quad 4\to A=4m,m\in Z$
Mà $12(3+...+3^{2019})\quad\vdots\quad 3, 4$ chia $3$ dư $1$
$\to A$ chia $3$ dư $1$
$\to A=3n+1,n\in Z$
$\to 4m=3n+1$
$\to m$ chia $3$ dư $1\to m=3k+1, k\in Z$
$\to 4(3k+1)=3n+1$
$\to 12k+4=3n+1$
$\to 3n=12k+3$
$\to n=4k+1$
$\to A=3(4k+1)=12k+3$ chia $12$ dư $3$
