Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
B = 1 + 3 + {3^2} + {3^3} + .... + {3^{2020}}\\
\Leftrightarrow 3B = 3.\left( {1 + 3 + {3^2} + {3^3} + ..... + {3^{2020}}} \right)\\
\Leftrightarrow 3B = 3 + {3^2} + {3^3} + {3^4} + ..... + {3^{2021}}\\
\Rightarrow 3B - B = \left( {3 + {3^2} + {3^3} + {3^4} + ..... + {3^{2021}}} \right) - \left( {1 + 3 + {3^2} + {3^3} + .... + {3^{2020}}} \right)\\
\Leftrightarrow 2B = {3^{2021}} - 1\\
2B{.3^x} + {3^x} = {3^{2025}}\\
\Leftrightarrow {3^x}.\left( {2B + 1} \right) = {3^{2025}}\\
\Leftrightarrow {3^x}{.3^{2021}} = {3^{2025}}\\
\Leftrightarrow {3^{x + 2021}} = {3^{2025}}\\
\Leftrightarrow x + 2021 = 2025\\
\Leftrightarrow x = 4\\
2,\\
C = 7 + {7^2} + {7^3} + .... + {7^{2020}}\\
\Leftrightarrow 7C = 7.\left( {7 + {7^2} + {7^3} + ..... + {7^{2020}}} \right)\\
\Leftrightarrow 7C = {7^2} + {7^3} + {7^4} + ..... + {7^{2021}}\\
\Rightarrow 7C - C = \left( {{7^2} + {7^3} + {7^4} + ..... + {7^{2021}}} \right) - \left( {7 + {7^2} + {7^3} + .... + {7^{2020}}} \right)\\
\Leftrightarrow 6C = {7^{2021}} - 7\\
{7^{x + 12}} - 7 = 6C\\
\Leftrightarrow {7^{x + 12}} - 7 = {7^{2021}} - 7\\
\Leftrightarrow {7^{x + 12}} = {7^{2021}}\\
\Leftrightarrow x + 12 = 2021\\
\Leftrightarrow x = 2009
\end{array}\)
