Cho S = 3 + 3 + 3+ + … + 32021. Chứng minh S không chia hết cho 13. 20/09/2021 Bởi Ariana Cho S = 3 + 3 + 3+ + … + 32021. Chứng minh S không chia hết cho 13.
Tham khảo ` S=3+3^2+3^3+…+3^{2021}` `⇒S=(3+3^2)+(3^3+3^4+3^5)+….+(3^{2019}+3^{2020}+3^{2021})` `⇒S=12+3^3.(1+3+3^2)+…+3^{2019}.(1+3+3^2)` `⇒S=12+(1+3+3^2).(3^3+…+3^{2019})` `⇒S=12+13.(3^3+…+3^{2019})` `⇒S` chia `13` dư `12` `⇒S` không chia hết `13` `\text{©CBT}` Bình luận
Tham khảo
` S=3+3^2+3^3+…+3^{2021}`
`⇒S=(3+3^2)+(3^3+3^4+3^5)+….+(3^{2019}+3^{2020}+3^{2021})`
`⇒S=12+3^3.(1+3+3^2)+…+3^{2019}.(1+3+3^2)`
`⇒S=12+(1+3+3^2).(3^3+…+3^{2019})`
`⇒S=12+13.(3^3+…+3^{2019})`
`⇒S` chia `13` dư `12`
`⇒S` không chia hết `13`
`\text{©CBT}`