\(3A=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow A=\dfrac{3^{100}-1}{2}\)
\(2\cdot A+3=3^n\)
\(\Leftrightarrow3^n=3^{100}\)
hay n=100
`=> 3A = 3 + 3^2 + ... + 3^100`
`=> 3A - A = (3+3^2 + ... + 3^100) - (1 + 3 + ... + 3^99)`
`=> 2A = 3^100 - 1`
`=> 2A + 3 = 3^100 + 2`
Mà `3^100 + 2 cancel vdots 3 =>` Vô lý