a)A=3+32+33+...+32004
=>3A=32+33+34+...+32005
=>3A-A=(32+33+34+...+32005)-(3+32+33+...+32004)
=>2A=32+33+34+...+32005-3-32-33-...-32004
=>2A=32005-3
=>A=0,10025
\(A=3+3^2+3^3+...+3^{2004}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2004}+3^{2005}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2004}+3^{2005}\right)-\left(3+3^2+3^3+...+3^{2004}\right)\)
\(\Rightarrow3A-A=2A=3^{2005}-3\)
\(\Rightarrow2A=3^{2005}-3\)
\(\Rightarrow A=\frac{3^{2005}-3}{2}\)
Vậy \(A=\frac{3^{2005}-3}{2}\)