cho B=2^0+2^2+2^3+….+3^60. Chứng tỏ C chia hết cho 3,7,15 02/07/2021 Bởi Clara cho B=2^0+2^2+2^3+….+3^60. Chứng tỏ C chia hết cho 3,7,15
sửa đề: Cho `B=2+2^2+2^3+….+2^60`. Chứng tỏ B chia hết cho $3,7,15$ Giải : `B=2+2^2+2^3+….+2^60` `=(2+2^2)+(2^3+2^4)+…+(2^59+2^60)` `=2(1+2)+2^3(1+2)+…+2^59(1+2)` `=> B vdots 3` (do `1+2=3 vdots 3`) `B=2+2^2+2^3+….+2^60` `=(2+2^2+2^3)+(2^4+2^5+2^6)+…+(2^58+2^59+2^60)` `=2(1+2+4)+2^4(1+2+4)+…+2^58(1+2+4)` `=> B vdots 7` (do `1+2+4=7 vdots 7`) `B=2+2^2+2^3+….+2^60` `=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+…+(2^57+2^58+2^59+2^60)` `=2(1+2+4+8)+2^4(1+2+4+8)+…+2^57(1+2+4+8)` `=> B vdots 15` (do `1+2+4+8=15 vdots 15`) Vậy `B vdots 3;7;15` Bình luận
sửa đề:
Cho `B=2+2^2+2^3+….+2^60`. Chứng tỏ B chia hết cho $3,7,15$
Giải :
`B=2+2^2+2^3+….+2^60`
`=(2+2^2)+(2^3+2^4)+…+(2^59+2^60)`
`=2(1+2)+2^3(1+2)+…+2^59(1+2)`
`=> B vdots 3` (do `1+2=3 vdots 3`)
`B=2+2^2+2^3+….+2^60`
`=(2+2^2+2^3)+(2^4+2^5+2^6)+…+(2^58+2^59+2^60)`
`=2(1+2+4)+2^4(1+2+4)+…+2^58(1+2+4)`
`=> B vdots 7` (do `1+2+4=7 vdots 7`)
`B=2+2^2+2^3+….+2^60`
`=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+…+(2^57+2^58+2^59+2^60)`
`=2(1+2+4+8)+2^4(1+2+4+8)+…+2^57(1+2+4+8)`
`=> B vdots 15` (do `1+2+4+8=15 vdots 15`)
Vậy `B vdots 3;7;15`