B=1+3^1+3^2+3^3+3^4+…+3^2019 chia cho 13 ;40 dư là ?

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`