Ta có:
$S = 1 - 4 + 4^2 - 4^3 + ... + 4^{98} - 4^{99}$
$a)$ $⇔ S = (1 - 4 + 4^2 - 4^3) + (4^4 - 4^5 + 4^6 - 4^7) + ... + (4^{96} - 4^{97} + 4^{98} - 4^{99})$
$⇔ S =(1 - 4 + 4^2 - 4^3) +4^4(1 - 4 + 4^2 - 4^3)+ .... + 4^{96} . (1 - 4 + 4^{2} - 4^3)$
$⇔ S = -51 + 4^4 . (-51) + ... + 4^{96} . (-51)$
$⇔ S = -51 . (1 +4^4 + ... + 4^{96}) \vdots -51$
$⇒ S \vdots -51$($đpcm$).
$b$) $S = 1 - 4 + 4^2 - 4^3 + ... + 4^{98} - 4^{99}$
$⇔ 4S = 4 - 4^2 + 4^3 - 4^4 + ... + 4^{99} - 4^{100}$
$⇔ 4S + S = (4 - 4^2 + 4^3 - 4^4 + ... + 4^{99} - 4^{100}) + (1 - 4 + 4^2 - 4^3 + ... + 4^{98} - 4^{99}$
$⇔ 5S = 1 - 4^{100}$
$⇔ S = \dfrac{1 - 4^{100}}{5}$
Vì $S ∈ Z$
$⇒ 1 - 4^{100} \vdots 5$
$⇒ 1 - 4^{100} = 5k$ ($k ∈ Z$)
$⇒ 4^{100} = 5k+1$ chia $5$ dư $1$
Vậy $4^{100}$ chia $5$ dư $1$.
