Chứng minh rằng: B = 3^4 + 3^5+ 3^6+ ... + 3^120 chia hết cho 13.

Ta có \[B = {3^4} + {3^5} + {3^6} + ... + {\rm{ }}{{\rm{3}}^{120}}\]                    

\[ = \left( {{3^4} + {3^5} + {3^6}} \right) + \left( {{3^7} + {3^8} + {3^9}} \right) + ... + \;\left( {{3^{118}} + {3^{119}} + {3^{120}}} \right)\]

\[ = {3^4} \cdot \left( {1 + 3 + {3^2}} \right) + {3^7} \cdot \left( {1 + 3 + {3^2}} \right) + ... + \;{3^{118}} \cdot \left( {1 + 3 + {3^2}} \right)\]

\( = \,{3^4} \cdot 13 + \,{3^7} \cdot 13 + ... + {3^{115}} \cdot 13 + {3^{118}} \cdot 13\)

\[\; = 13.\left( {{3^4} + {3^7} +  \ldots  + {3^{115}} + {3^{118}}} \right)\].

Vì \(13\,\, \vdots \,\,13\) nên \(\;13 \cdot \left( {{3^4} + {3^7} +  \ldots  + {3^{115}} + {3^{118}}} \right)\,\, \vdots \,\,13\).

Vậy \[B = {3^4} + {3^5} + {3^6} + ... + {\rm{ }}{{\rm{3}}^{120}}\] chia hết cho 13.