Toán 2 + 2 mũ 2 + 2 mũ 3 + …. + 2 mũ 100 chứng tỏ s+5 chia hết cho7 20/09/2021 By Julia 2 + 2 mũ 2 + 2 mũ 3 + …. + 2 mũ 100 chứng tỏ s+5 chia hết cho7
Tham khảo Đặt `S=2+2^2+2^3+2^4+….+2^{100}` `⇒S+5=(2+5)+(2^2+2^3+2^4)+…+(2^{98}+2^{99}+2^{100})` `⇒S+5=7+2.(2+2^2+2^3)+….+2^{97}.(2+2^2+2^3)` `⇒S+5=7+(2+2^2+2^3).(2+…+2^{97})` `⇒S+5=7+14.(2+..+2^{97})` Vì `7 \vdots 7,14 \vdots 7⇒14.(2+..+2^{97}) \vdots 7` Do đó `7+14.(2+..+2^{97}) \vdots 7` hay `S+5 \vdots 7` `\text{©CBT}` Trả lời
Tham khảo
Đặt `S=2+2^2+2^3+2^4+….+2^{100}`
`⇒S+5=(2+5)+(2^2+2^3+2^4)+…+(2^{98}+2^{99}+2^{100})`
`⇒S+5=7+2.(2+2^2+2^3)+….+2^{97}.(2+2^2+2^3)`
`⇒S+5=7+(2+2^2+2^3).(2+…+2^{97})`
`⇒S+5=7+14.(2+..+2^{97})`
Vì `7 \vdots 7,14 \vdots 7⇒14.(2+..+2^{97}) \vdots 7`
Do đó `7+14.(2+..+2^{97}) \vdots 7`
hay `S+5 \vdots 7`
`\text{©CBT}`