Cho \(A = 1 + 3 + {3^2} + {3^3} + ... + {3^{101}}\). Chứng minh \(A\,\, \vdots \,\,13\).

Ta có \(A = 1 + 3 + {3^2} + {3^3} + ... + {3^{101}}\)

\[ = \left( {1 + 3 + {3^2}} \right) + \left( {{3^3} + {3^4} + {3^5}} \right) + ... + \left( {{3^{99}} + {3^{100}} + {3^{101}}} \right)\]

\[ = \left( {1 + 3 + {3^2}} \right) + {3^3}\left( {1 + 3 + {3^2}} \right) + ... + {3^{99}}\left( {1 + 3 + {3^2}} \right)\]

\[ = \left( {1 + 3 + {3^2}} \right)\left( {1 + {3^3} + ... + {3^{99}}} \right)\]

\( = 13 \cdot \left( {1 + {3^3} + ... + {3^{99}}} \right)\)

Vì \(13\,\, \vdots \,\,13\) nên \(13 \cdot \left( {1 + {3^3} + ... + {3^{99}}} \right)\,\, \vdots \,\,13\) hay \(A\,\, \vdots \,\,13\) (đpcm).