Đáp án:
$\begin{array}{l}
3 + {3^3} + {3^5} + ... + {3^{2n + 1}}\\
= \left( {3 + {3^3}} \right) + \left( {{3^5} + {3^7}} \right) + ... + \left( {{3^{2n - 1}} + {3^{2n + 1}}} \right)\\
= 3\left( {1 + {3^2}} \right) + {3^5}\left( {1 + {3^2}} \right) + ... + {3^{2n - 1}}\left( {1 + {3^2}} \right)\\
= 3.10 + {3^3}.10 + .. + {3^{2n - 1}}.10\\
= 10.\left( {3 + {3^3} + .. + {3^{2n - 1}}} \right) \vdots 10\\
mà\,3 + {3^3} + {3^5} + ... + {3^{2n + 1}} \vdots 3\\
\Rightarrow 3 + {3^3} + {3^5} + ... + {3^{2n + 1}} \vdots 3.10 = 30\\
Vậy\,3 + {3^3} + {3^5} + ... + {3^{2n + 1}} \vdots 30
\end{array}$
