Đáp án: $7$
Giải thích các bước giải:
Ta có:
$A=1+3+3^1+3^2+...+3^{2020}+3^{2021}$
$\to A=1+3+(3^1+3^2)+(3^3+3^4)+...+(3^{2019}+3^{2020})+3^{2021}$
$\to A=1+3+(3+3^2)+3^2(3+3^2)+...+3^{2018}(3+3^{2})+3^{2021}$
$\to A=1+3+(3+3^2)(1+3^2+...+3^{2018})+3^{2021}$
$\to A=4+12(1+3^2+...+3^{2018})+3^{2021}$
Ta có: $3\equiv (-1)(mod 4)\to 3^{2021}\equiv -1(mod 4)$
$\to A\equiv -1(mod 4)\to A=4k-1, k\in N$
Lại có: $4\equiv 1(mod 3)\to A=3q+1, q\in N$
$\to 4k-1=3q+1$
$\to 3q+2=4k$
$\to k$ chia $3$ dư $2\to k=3m+2$
$\to 3q+2=4(3m+2)=12m+8$
$\to 3q=12m+6$
$\to q=4m+2$
$\to A=3\cdot (4m+2)+1=12m+7$
$\to A$ chia $12$ dư $7$
