\(3A=3+3^2+3^3+...+3^{2025}\)
\(3A-A=3+3^2+3^3+...+3^{2025}-\left(1+3+3^2+...+3^{2024}\right)=-1+3^{2025}\)
\(A=\dfrac{-1+3^{2025}}{2}\)
\(A=1+3+3^2+...+3^{2024}\\ 3A=3+3^2+3^3+...+3^{2025}\\ 3A-A=\left(3+3^2+3^3+...+3^{2025}\right)-\left(1+3+3^2+...+3^{2024}\right)\\ 2A=3^{2025}-1\\ A=\dfrac{3^{2025}-1}{2}\)
\(A=1+3+3^2+...+3^{2024}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2025}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2025}\right)-\left(1+3+3^2+...+3^{2024}\right)\)
\(\Rightarrow2A=3^{2025}-1\)
\(\Rightarrow A=\dfrac{3^{2025}-1}{2}\)
Vậy \(A=\dfrac{3^{2025}-1}{2}\)
