Đáp án:
$\begin{array}{l}
1)\\
a)B = 1 + 5 + {5^2} + {5^3} + ... + {5^{100}}\\
\Rightarrow 5.B = 5 + {5^2} + {5^3} + ... + {5^{101}}\\
\Rightarrow 5B - B = 4B = {5^{101}} - 1\\
\Rightarrow B = \dfrac{{{5^{101}} - 1}}{4}\\
b)D = 1 + 9 + {9^2} + ... + {9^{2019}}\\
\Rightarrow 9D = 9 + {9^2} + {9^3} + .. + {9^{2020}}\\
\Rightarrow 8D = {9^{2020}} - 1\\
\Rightarrow D = \dfrac{{{9^{2020}} - 1}}{8}\\
B2)\\
a)A = 2 + {2^2} + {2^3} + ... + {2^{60}}\\
= \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right) + ... + \left( {{2^{59}} + {2^{60}}} \right)\\
= 2\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + ... + {2^{59}}\left( {1 + 2} \right)\\
= 2.3 + {2^3}.3 + ... + {2^{59}}.3\\
= \left( {2 + {2^3} + ... + {2^{59}}} \right).3 \vdots 3\\
A = \left( {2 + {2^3}} \right) + \left( {{2^2} + {2^4}} \right) + ... + \left( {{2^{58}} + {2^{60}}} \right)\\
= 2\left( {1 + {2^2}} \right) + {2^2}.\left( {1 + {2^2}} \right) + ... + {2^{58}}\left( {1 + {2^2}} \right)\\
= 2.5 + {2^2}.5 + ... + {2^{58}}.5\\
= \left( {2 + {2^2} + {2^5} + ... + {2^{58}}} \right).5 \vdots 5\\
A = \left( {2 + {2^2} + {2^3}} \right) + ... + \left( {{2^{58}} + {2^{59}} + {2^{60}}} \right)\\
= 2\left( {1 + 2 + {2^2}} \right) + .. + {2^{58}}\left( {1 + 2 + {2^2}} \right)\\
= 2.7 + {2^4}.7 + ... + {2^{58}}.7\\
= \left( {2 + {2^4} + ... + {2^{58}}} \right).7 \vdots 7\\
b)B = 7 + {7^2} + {7^3} + ... + {7^{99}} + {7^{100}}\\
= 7\left( {1 + 7} \right) + {7^3}.\left( {1 + 7} \right) + ... + {7^{99}}\left( {1 + 7} \right)\\
= \left( {7 + {7^3} + ... + {7^{99}}} \right).8 \vdots 8\\
B = \left( {7 + {7^3}} \right) + \left( {{7^2} + {7^4}} \right) + ... + \left( {{7^{98}} + {7^{100}}} \right)\\
= 7\left( {1 + {7^2}} \right) + {7^2}\left( {1 + {7^2}} \right) + ... + {7^{98}}\left( {1 + {7^2}} \right)\\
= \left( {7 + {7^2} + {7^5} + {7^6} + ... + {7^{98}}} \right).50 \vdots 50
\end{array}$
