1) Cho A = 2 mũ 1 + 2² + 2³ + 2 mũ 4 +….+ 2 mũ 59 + 2 mũ 60 a. Chứng minh rằng A chia hết cho 7 b. Chứng minh rằng A chia hết cho 15

$A = 2^1 + 2^2 + 2^3 +2^4 +\dots + 2^{59} + 2^{60}$

Ta có:

$A = (2^1 +2^2 + 2^3) + (2^4 + 2^5 + 2^6) + \dots + (2^{58} + 2^{59} + 2^{60})$

$A = (2^1 + 2^2 + 2^3) + 2^3.(2^1 + 2^2 + 2^3) + \dots + 2^{57}.(2^1 + 2^2 + 2^3)$

$A = (2^1 + 2^2 + 2^3)(2^3 + 2^6 + \dots + 2^{57})$

$A = (2 + 4 + 8)(2^3 + 2^6 + \dots + 2^{57})$

$A = 14.(2^3 + 2^6 + \dots + 2^{57})$

Do $14 \,\,\,\vdots\,\,\,7$

nên $14.(2^3 + 2^6 + \dots + 2^{57})\,\,\,\vdots\,\,\,7$

hay $A \,\,\,\vdots\,\,\,7$

Ta có: $A = (2^1 +2^2 + 2^3 + 2^4) + (2^5 + 2^6 + 2^7 + 2^8) + \dots + (2^{57} +2^{58} + 2^{59} + 2^{60})$

$A = (2^1 +2^2 + 2^3 + 2^4) + 2^4.(2^1 +2^2 + 2^3 + 2^4) + \dots + 2^{56}.(2^1 +2^2 + 2^3 + 2^4)$

$A = (2^1 +2^2 + 2^3 + 2^4)(2^4 + 2^8 + \dots + 2^{56})$

$A = (2 + 4 + 8 + 16).(2^4 + 2^8 + \dots + 2^{56})$

$A = 30(2^4 + 2^8 + \dots + 2^{56})$

Do $30 \,\,\,\vdots\,\,\,15$

nên $30(2^4 + 2^8 + \dots + 2^{56})\,\,\,\vdots\,\,\,15$

hay $A\,\,\,\vdots\,\,\,15$