Question

[Phạm Thị Xuân Quỳnh]

Chứng minh rằng A=2+2²+2³+...+2⁶⁰ Chia hết cho 3, 7 và 15

Answer 1

[Akai Haruma]

Lời giải:

CM $A\vdots 7$:

$A=(2+2^2+2^3)+(2^4+2^5+2^6)+....+(2^{58}+2^{59}+2^{60})$

$=2(1+2+2^2)+2^4(1+2+2^2)+....+2^{58}(1+2+2^2)$

$=(1+2+2^2)(2+2^4+....+2^{58})$

$=7(2+2^4+....+2^{58})\vdots 7$

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CM $A\vdots 3$:

$A=(2+2^2)+(2^3+2^4)+....+(2^{59}+2^{60})$

$=2(1+2)+2^3(1+2)+....+2^{59}(1+2)$

$=(1+2)(2+2^3+...+2^{59})=3(2+2^3+....+2^{59})\vdots 3$

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CM $A\vdots 15$:

$A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+...+(2^{57}+2^{58}+2^{59}+2^{60})$

$=2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+....+2^{57}(1+2+2^2+2^3)$

$=(1+2+2^2+2^3)(2+2^5+...+2^{57})$

$=15(2+2^5+...+2^{57})\vdots 15$