A = 3 + 32 + 33 + ... + 3100
3A = 3 ( 3 + 32 + 33 + ... + 3100)
= 32 + 33 + ... + 3101
3A - A = ( 32 + 33 + ... + 3100) - 3 + 32 + 33 + ... + 3100
2A = 3101 - A
\(\Rightarrow\) 2A + 3 = 3101 - 3 + 3 = 3101
mà 2A + 3 = 3n \(\Rightarrow\)3n = 3 101\(\Rightarrow\)n = 101
\(A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow2A+3=3^{101}-3+3=3^{101}=3^n\)
\(\Rightarrow n=101\)
\(A=3+3^2+3^3+...+3^{99}+3^{100}\)
\(3A=3^2+3^3+....+3^{100}+3^{101}\)
\(\Rightarrow3A-A=\left(3^2+3^3+....+3^{100}+3^{101}\right)-\left(3+3^2+3^3+...+3^{99}+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(2A+3=3^{101}\)
\(\Rightarrow3^n=3^{101}\)
\(\Rightarrow n=101\)
A=3+3^2+3^3+...........3^100
3A=3^2+3^3+3^4+.........+3^101
3A-A=(3^2+3^3+......+3^101)-(3+3^2+....+3^100)
2A=3^101-3
A=(3^101-3):2
thay vào biểu thức ta có :(3^101-3):2.2+3=3^n
3^101=3^n
n=101
vậy n=101
\(A=3+3^2+3^3+......+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+.......+3^{100}+3^{101}\)
\(\Rightarrow3A-A=2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Ta có: \(2A+3=3^n\)
\(\Rightarrow2.\frac{3^{101}-3}{2}+3=3^n\)
\(\Rightarrow3^{101}-3+3=3^n\)
\(\Rightarrow3^{101}=3^n\)
\(\Rightarrow n=101\)
Vậy n = 101
