\(A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow2A=\frac{\left(3^{101}-3\right)}{2}\)
Ta có:
\(A=3+3^2+3^3+........+3^{100}\Rightarrow3A=3^2+3^3+.......+3^{101}\Rightarrow3A-A=2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}=3n\Rightarrow n=3^{100}\)
\(A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+..+3^{100}\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}-3+3=3^{101}\)
\(\Rightarrow n=101\)
Ta có A=3+32+...+3100
=> 3A=32+33+...+3101
=>2A=3A - A=3101-3 (1)
Ta lại có
2A+3 =3n (ở đề hình như bị ghi sai !!!!!!!!)
Từ (1) =>3101-3+3=3n
=>3101=3n
=>n=101
Good luck !!!!!!!!!!!!!!!!!! :)