\(A=1+3+3^2+3^3+3^{2000}\)
\(\Rightarrow3A=3\left(1+3+3^2+3^3+...+3^{2000}\right)\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2000}+3^{2001}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2001}\right)-\left(1+3+3^2+3^3+...+3^{2000}\right)\)
\(\Rightarrow2A=3+3^2+3^3+...+3^{2001}-1-3-3^2-3^3-...-3^{2000}\)
\(\Rightarrow2A=3^{2001}-1\)
\(\Rightarrow n=2001\)
Vậy \(n=2001\)
Theo bài ra, ta có:
A = 1 + 3 + 32 + .....+ 32000
\(\Rightarrow\) 3A = 3(1 + 3 + 32 + .......+ 32000)
\(\Rightarrow\) 3A = 3 + 32 + 33 + ....... + 32001
\(\Rightarrow\)3A-A = (3 + 32 + 33 + ....... + 32001) - (1 + 3 + 32 + .......+ 32000)
\(\Rightarrow\) 2A = 32001 - 1
Ta lại có: 2A = 3n - 1
= 32001 - 1
\(\Rightarrow\) n = 2001
Vậy n = 2001
