Đặt A = \(3+3^{^{ }2}+3^{^{ }3}+....+3^{^{ }32}\)
=>3A=3.(\(3^1+3^2+3^3+.....+3^{^{ }32}\))
=>3A=\(3^2+3^3+3^4+...+3^{33}\)
=>3A-A = (\(3^2+3^3+3^{^{ }4}+.....+3^{33}\))-(\(3^1+3^2+3^3+....+3^{32}\))
=>2A=\(3^{33}-3^1\)
=>A = \(\frac{3^{33}-3}{2}\)