Đáp án:
$\begin{array}{l}
C = 1 + 3 + {3^2} + {3^3} + ... + {3^{2003}}\\
\Rightarrow 3.C = 3 + {3^2} + {3^3} + {3^4} + ... + {3^{2004}}\\
\Rightarrow 3C - C = \left( {3 + {3^2} + {3^3} + {3^4} + ... + {3^{2004}}} \right)\\
- \left( {1 + 3 + {3^2} + {3^3} + ... + {3^{2003}}} \right)\\
\Rightarrow 2C = {3^{2004}} - 1\\
\Rightarrow C = \dfrac{{{3^{2004}} - 1}}{2} < {3^{2003}}\\
B = {3^{2004}} \Rightarrow \dfrac{B}{3} = {3^{2003}}\\
\Rightarrow C < \dfrac{B}{3}
\end{array}$
