\(A=3+3^2+3^3+...+3^{100}\)
\(3A=3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(2A=3^{101}-3\)
\(2A+3=2^{101}-3+3\)
\(2A+3=2^{101}\)
=> 3A = 32 + 33 + 34 +...+ 3101 => 3A -A = 32 + 33 + 34 +...+ 3101 - (3 + 32 + 33 +...+ 3100) => 2A = 32 + 33 + 34 +...+ 3101 - 3 - 32 - 33 -...- 3100 2A = 3101 - 3 2A + 3 = 3101 => ĐPCM
Ta có: A = 3 + 32 + 33 + ... + 3100
=> 3A = 32 + 33 + 34 + ... + 3101
=> 3A - A = 3101 - 3
=> 2A = 3101 - 3
=> 2A + 3 = 3101
Vậy 2a + 3 là 1 lũy thừa của 3