\(B=4+3^2+3^3+...+3^{2004}\)
\(\Rightarrow B=1+3+3^2+3^3+...+3^{2004}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{2005}\)
\(\Rightarrow3B-B=3+3^2+3^3+...+3^{2005}-1-3-3^2-...-3^{2004}\)
\(\Rightarrow2B=3^{2005}-1\)
\(\Rightarrow B=\frac{3^{2005}-1}{2}< \frac{3^{2005}}{2}< 3^{2005}=C\)
Vậy B < C