`S=1+2+2^2+2^3+...+2^25` Chứng minh S chia hết cho `7;15;21;35`

Đáp án+Giải thích các bước giải:

$S=1+2+2^2+2^3+...+2^{25}\\=(1+2+2^2)+. . . +2^{23}.(1+2+2^2)\\=7+. . . +7.2^{23}\\=7.(1+. . .2^{23}) \vdots7$

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$S=1+2+2^2+2^3+...+2^{25}\\=(1+2+2^2+2^3)+. . .+2^{22}.(1+2+2^2+2^3)\\=15+. . . +15.2^{22}\\=15.(1+. . . 2^{22})\vdots 15$

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$S=1+2+2^2+2^3+...+2^{25}\\=(1+2^2+2^4)+. . . +2^{19}.(1+2^2+2^4)\\=21+. . . +21.2^{19}\\=21.(1+. . . +2^{19})\vdots 21$

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$S=1+2+2^2+2^3+...+2^{25}\\=(1+2+2^5)+. . . +2^{19}.(1+2+2^5)\\=35+. . . +35.2^{19}\\=35.(1+. . . +2^{19})\vdots 35$

Đáp án `+` Lời giải:

Để `S\vdots7`

`S=1+2+2^2+2^3+...+2^25`

`->S=(1+2+2^2)+(2^3+2^4+2^5)+...+(2^23+2^24+2^25)`

`->S=(1+2+4)+2^3 . (1+2+2^2)+...+2^23 . (1+2+2^2)`

`->S=7+2^3 . (1+2+4)+...+2^23 . (1+2+4)`

`->S=7+2^3 . 7+...+2^23 . 7`

`->S=7 . (1+2^3+...+2^23)`

`->S\vdots7(1)`

Để `S\vdots15`

`S=1+2+2^2+2^3+...+2^25`

`->S=(1+2+2^2+2^3)+(2^4+2^5+2^6+2^7)+...+(2^22+2^23+2^24+2^25)`

`->S=(1+2+4+8)+2^4 . (1+2+2^2+2^3)+...+2^22 . (1+2+2^3+2^4)`

`->S=15+2^4 . (1+2+4+8)+...+2^22 . (1+2+4+8)`

`->S=15+2^4 . 15+..+2^22 . 15`

`->S=15 . (1+2^4+...+2^22)`

`->S\vdots15(2)`

Để `S\vdots21`

`S=1+2+2^2+2^3+...+2^25`

`->S=(1+2^2+2^4)+(2^5+2^7+2^9)+...+(2^19+2^21+2^23)`

`->S=(1+4+16)+2^5 . (1+2^2+2^4)+...+2^19 . (1+2^2+2^4)`

`->S=21+2^5 . (1+4+16)+...+2^19 . (1+4+16)`

`->S=21+2^5 . 21+..+2^19 . 21`

`->S=21 . (1+2^5+...+2^19)`

`->S\vdots21(3)`

Để `S\vdots35`

`S=1+2+2^2+2^3+...+2^25`

`->S=(1+2+2^5)+(2^4+2^5+2^9)+...+(2^19+2^20+2^24)`

`->S=(1+2+32)+2^4 . (1+2+2^5)+...+2^19 . (1+2+2^5)`

`->S=35+2^4 . (1+2+32)+...+2^19 . (1+2+32)`

`->S=35+2^4 . 35+..+2^19 . 35`

`->S=35 . (1+2^4+...+2^19)`

`->S\vdots35(4)`

Từ `(1), (2), (3), (4)` thì `S` đều chia hết cho `7; 15; 21; 35` và đề bài đã được chứng minh.