Đáp án:
Giải thích các bước giải:
`A=2+2^2+ 2^3+ .... +2^58+ 2^59+ 2^60`
`+)` Với `A\vdots3`
`=>A=(2+2^2)+(2^3+2^4)+...+(2^57+2^58)+(2^59+2^60)`
`=2(1+2)+2^3(1+2)+...+2^57(1+2)+2^59(1+2)`
`=2.3+2^3. 3+...+2^57 .3+2^59 .3`
`=3.(2+2^3+....+2^57+2^59)`
Vì `3\vdots3`
`=>3.(2+2^3+....+2^57+2^59)\vdots3`
`=>A\vdots3(1)`
`+)` Với `A\vdots7`
`=>A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^55+2^56+2^57)+(2^58+2^59+2^60)`
`=2.(1+2+2^2)+2^4.(1+2+2^2)+...+2^58.(1+2+2^2)`
`=2.7+2^4. 7+...+2^55 .7+2^58 .7`
`=7.(2+2^4+....+2^55+2^58)`
Vì `7\vdots7`
`=>7.(2+2^4+....+2^55+2^58)\vdots7`
`=>A\vdots7(2)`
`+)` Với `A\vdots15`
`=>A=(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+2^2+2^3)+2^5.(1+2+2^2+2^3)+...+2^57.(1+2+2^2+2^3)`
`=2.15+2^5. 15+...+2^57 .15` `=15.(2+2^5+....+2^57)`
Vì `15\vdots15`
`=>15.(2+2^5+....+2^57)\vdots15`
`=>A\vdots15(3)`
Từ `(1)(2)(3)=>dpcm`
