Đáp án + Giải thích các bước giải:
`B=1+3^{1}+3^{2}+3^{3}+3^{4}+....+3^{2019}`
`=(1+3^{1}+3^{2})+(3^{3}+3^{4}+3^{5})+....+(3^{2017}+3^{2018}+3^{2019})`
`=(1+3+3^{2})+3^{3}(1+3+3^{2})+....+3^{2017}(1+3+3^{2})`
`=13+3^{3}.13+....+3^{2017}.13`
`=13(1+3^{3}+...+3^{2017})` `\vdots 13`
`--------------`
`B=1+3^{1}+3^{2}+3^{3}+3^{4}+....+3^{2019}`
`=(1+3^{1}+3^{2}+3^{3})+....+(3^{2016}+3^{2017}+3^{2018}+3^{2019})`
`=(1+3+3^{2}+3^{3})+....+3^{2016}(1+3+3^{2}+3^{3})`
`=40+...+3^{2016}.40`
`=40.(1+...+3^{2016})` `\vdots 40`
Vậy số dư khi chia `B` cho `13` là : `0`
số dư khi chia `B` cho `40` là : `0`
