Đáp án:
$\begin{array}{l}
a)A = {7^0} + {7^1} + {7^2} + ... + {7^{2018}} + {7^{2019}} + {7^{2020}}\\
\Rightarrow 7A = 7 + {7^2} + {7^3} + ... + {7^{2019}} + {7^{2020}} + {7^{2021}}\\
\Rightarrow 7A - A = {7^{2021}} - {7^0}\\
\Rightarrow 6A = {7^{2021}} - 1\\
\Rightarrow A = \frac{{{7^{2021}} - 1}}{6}\\
b)A = {7^0} + {7^1} + {7^2} + ... + {7^{2018}} + {7^{2019}} + {7^{2020}}\\
= 1 + \left( {{7^1} + {7^2}} \right) + \left( {{7^3} + {7^4}} \right) + ... + \left( {{7^{2019}} + {7^{2020}}} \right)\\
= 1 + 7\left( {1 + 7} \right) + {7^3}\left( {1 + 7} \right) + .... + {7^{2019}}\left( {1 + 7} \right)\\
= 1 + 7.8 + {7^3}.8 + ... + {7^{2019}}.8\\
= 1 + \left( {7 + {7^3} + ... + {7^{2019}}} \right).8\\
Do:\left( {7 + {7^3} + ... + {7^{2019}}} \right).8 \vdots 8\\
\Rightarrow 1 + \left( {7 + {7^3} + ... + {7^{2019}}} \right).8\,chia\,8\,dư\,1
\end{array}$
Vậy A chia 8 dư 1
