Đáp án:
$\begin{array}{l}
{3^1} + {3^2} + {3^3} + ... + {3^{2010}}\\
= \left( {{3^1} + {3^2}} \right) + \left( {{3^3} + {3^4}} \right) + ... + \left( {{3^{2009}} + {3^{2010}}} \right)\\
= 3\left( {1 + 3} \right) + {3^3}\left( {1 + 3} \right) + ... + {3^{2009}}\left( {1 + 3} \right)\\
= 3.4 + {3^3}.4 + ... + {3^{2009}}.4\\
= 4\left( {3 + {3^3} + ... + {3^{2009}}} \right) \vdots 4
\end{array}$
Vậy ${3^1} + {3^2} + {3^3} + ... + {3^{2010}} \vdots 4$
