Ta có:A=3^0+3^1+3^2+...+3^2017
3A=3.(3^0+3^1+3^2+...+3^2017)
3A=3^1+3^2+3^3+...+3^2018
3A-A=(3^1+3^2+3^3+...+3^2018)-(3^0+3^1+3^2+...+3^2017)
2A=3^2018-3^0
2A=3^2018-1
2A+1=3^2018
Vì 3^2018 chia hết cho 3
=>2A+1 chia hết cho 3
=>2A chia 3 dư 2
=>A chia 3 dư 1
Vậy A chia 3 dư 1