\(A=1+3+3^2+....+3^{20}\)
\(\Leftrightarrow3A=3+3^2+...+3^{21}\)
\(\Leftrightarrow3A-A=\left(3+3^2+...+3^{21}\right)-\left(1+3+....+3^{20}\right)\)
\(\Leftrightarrow2A=3^{21}-1\)
\(\Leftrightarrow A=\dfrac{3^{21}-1}{2}\)
Mà \(B=3^{21}-1\)
\(\Leftrightarrow A< B\)
A=1+3+3^2+........+3^20
3A = 3 . ( 1 + 3 + \(3^2+...+3^{20}\))
3A = 3 + \(3^2+3^3+...+3^{21}\)
=> 3A - A = ( 3 + \(3^2+3^3+...+3^{21}\)) - ( \(1+3+3^2+3^{20}\) )
2A = \(3+3^2+3^3+...+3^{21}-1+3+3^2+...+3^{20}\)
=> A = \(\dfrac{3^{21}-1}{2}\)
Vì \(3^{21}-1\) > \(\dfrac{3^{21}-1}{2}\) nên => A < B
Vậy A < B
