Chứng minh `A=75.(4^2018+…+4^2+4) \vdots 4^2019 ` 09/11/2021 Bởi Harper Chứng minh `A=75.(4^2018+…+4^2+4) \vdots 4^2019 `
$A$ = $75$.($4^{2018}$ + … + $4^{2}$ + $4 $) = $75$.($4^{2018}$.$4^{2019}$+….+$4^{2}$.$4^{2019}$+$4^{2019}$). $\frac{1}{4^{2019}}$ = $75$.$4^{2019}$.($4^{2018}$ + … + $4^{2}$ + $4 $).$\frac{1}{4^{2019}}$ mà $4^{2019}$ ⋮ $4^{2019}$ ⇒ $A$ ⋮ $4^{2019}$ Vậy … Bình luận
$A$ = $75$.($4^{2018}$ + … + $4^{2}$ + $4 $)
= $75$.($4^{2018}$.$4^{2019}$+….+$4^{2}$.$4^{2019}$+$4^{2019}$). $\frac{1}{4^{2019}}$
= $75$.$4^{2019}$.($4^{2018}$ + … + $4^{2}$ + $4 $).$\frac{1}{4^{2019}}$
mà $4^{2019}$ ⋮ $4^{2019}$
⇒ $A$ ⋮ $4^{2019}$
Vậy …