\(A=3+3^2+3^3+.........+3^{100}\)
\(\Leftrightarrow3A=3^2+3^3+.........+3^{100}+3^{101}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+.....+3^{101}\right)-\left(3+3^2+......+3^{100}\right)\)
\(\Leftrightarrow2A=3^{101}-3\)
\(\Leftrightarrow2A+3=3^{101}\)
Mà \(2A+3=3^n\)
\(\Leftrightarrow3^{101}=3^n\)
\(\Leftrightarrow n=101\)
Vậy ..
A = 3 + 32 + 33 + ... + 3100
\(\Rightarrow\) 3A = 32 + 33 + 34 + ... + 3101
\(\Rightarrow\) 3A - A = (32 + 33 + 34 + ... + 3101) - (3 + 32 + 33 + ... + 3100)
\(\Rightarrow\) 2A = 3101 - 3
\(\Rightarrow\) 2A + 3 = 3101
\(\Rightarrow\) 3101 = 3n
\(\Rightarrow\) n = 101