Cho A = 1 + 3 + 3^2 + 3^3 + + 3^2021, B = 2^2022 : 2. Tính B - A

\(\begin{array}{l}A = 1 + 3 + {3^2} + {3^3} + ... + {3^{2021}}.\\3A = 3\left( {1 + 3 + {3^2} + {3^3} + ... + {3^{2021}}} \right) = 3 + {3^2} + ... + {3^{2022}}\\3A - A = 2A = {3^{2022}} - 1\\ \Rightarrow A = \frac{{{3^{2022}} - 1}}{2}\end{array}\)             \[\]

\(B - A = \frac{{{3^{2022}}}}{2} - \frac{{{3^{2022}} - 1}}{2} = \frac{{{3^{2022}} - {3^{2022}} + 1}}{2} = \frac{1}{2}\).