$A=1+7+{{7}^{2}}+{{7}^{3}}+...+{{7}^{2020}}$ . Hỏi A chia cho 57 dư bao nhiêu
$A={{7}^{0}}+{{7}^{1}}+{{7}^{2}}+{{7}^{3}}+...+{{7}^{2020}}$
Số số hạng là $\left( 2020-0 \right):1+1=2021$
Vì 2021 chia 3 dư 2 nên ta có
$A=\left( 1+7 \right)+\left( {{7}^{2}}+{{7}^{3}}+{{7}^{4}} \right)+...+\left( {{7}^{2018}}+{{7}^{2019}}+{{7}^{2020}} \right)$
$A=8+{{7}^{2}}\left( 1+7+{{7}^{2}} \right)+...+{{7}^{2018}}.\left( 1+7+{{7}^{2}} \right)$
$A=8+{{57.7}^{2}}+...+{{57.7}^{2018}}$
$A=8+57.\left( {{7}^{2}}+{{7}^{5}}+...+{{7}^{2018}} \right)$
Do đó $A$ chia 57 dư 8