Bài 1 :
$A = 4(1 + 4 + 4^2) + 4^4(1 + 4 + 4^2) +... + 4^{2018}(1 + 4 + 4^2)$
$⇔ A = (1 + 4 + 4^2).(4 + 4^4 + ... + 4^{2018})$
$⇔ A = 21. (4 + 4^4 + ... + 4^{2018}) ⋮ 21$
Bài 2 :
$A = (2 + 2^2) + (2^3 + 2^4) + (2^5 + 2^6) + ... + (2^{2009} + 2^{2010})$
$⇔A = 1(2 + 2^2) + 2^2(2 + 2^2) + ...+ 2^{2008}(2 + 2^{2010})$
$⇔ A = (2 + 2^2)(1 + 2 + ... + 2^{2008})$
$⇔ A = 6.(1 + 2 + ... + 2^{2008}) ⋮ 3$
$A = 2(1 + 2 + 2^2) + 2^4(1 + 2 + 2^2) + ... + 2^{2008}(1 + 2 + 2^2)$
$⇔A = 7.(2 + 2^4 + ... + 2^{2008}) ⋮ 7$
Vậy A chia hết cho 3, chia hết cho 7