Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3,\\
a,\\
S = 1 + 2 + {2^2} + .... + {2^{2021}}\\
\Leftrightarrow 2S = 2.\left( {1 + 2 + {2^2} + .... + {2^{2021}}} \right)\\
\Leftrightarrow 2S = 2 + {2^2} + {2^3} + ..... + {2^{2022}}\\
\Rightarrow 2S - S = \left( {2 + {2^2} + {2^3} + ..... + {2^{2022}}} \right) - \left( {1 + 2 + {2^2} + .... + {2^{2021}}} \right)\\
\Leftrightarrow S = {2^{2022}} - 1\\
b,\\
A = 1 + 3 + {3^3} + {3^3} + .... + {3^{2020}}\\
\Leftrightarrow 3A = 3.\left( {1 + 3 + {3^3} + {3^3} + .... + {3^{2020}}} \right)\\
\Leftrightarrow 3A = 3 + {3^2} + {3^3} + {3^4} + ..... + {3^{2021}}\\
\Rightarrow 3A - A = \left( {3 + {3^2} + {3^3} + {3^4} + ..... + {3^{2021}}} \right) - \left( {1 + 3 + {3^3} + {3^3} + .... + {3^{2020}}} \right)\\
\Leftrightarrow 2A = {3^{2021}} - 1\\
\Leftrightarrow A = \dfrac{{{3^{2021}} - 1}}{2}\\
c,\\
B = 4 + {4^2} + {4^3} + ..... + {4^{2020}}\\
\Leftrightarrow 4B = 4.\left( {4 + {4^2} + {4^3} + ..... + {4^{2020}}} \right)\\
\Leftrightarrow 4B = {4^2} + {4^3} + {4^4} + ..... + {4^{2021}}\\
\Leftrightarrow 4B - B = \left( {{4^2} + {4^3} + {4^4} + ..... + {4^{2021}}} \right) - \left( {4 + {4^2} + {4^3} + ..... + {4^{2020}}} \right)\\
\Leftrightarrow 3B = {4^{2021}} - 4\\
\Leftrightarrow B = \dfrac{{{4^{2021}} - 4}}{3}\\
d,\\
C = 5 + {5^2} + {5^3} + .... + {5^{2020}}\\
\Leftrightarrow 5C = 5.\left( {5 + {5^2} + {5^3} + .... + {5^{2020}}} \right)\\
\Leftrightarrow 5C = {5^2} + {5^3} + {5^4} + ..... + {5^{2021}}\\
\Leftrightarrow 5C - C = \left( {{5^2} + {5^3} + {5^4} + ..... + {5^{2021}}} \right) - \left( {5 + {5^2} + {5^3} + .... + {5^{2020}}} \right)\\
\Leftrightarrow 4C = {5^{2021}} - 5\\
\Leftrightarrow C = \dfrac{{{5^{2021}} - 5}}{4}
\end{array}\)
