Chứng tỏ rằng: A = 1 + 4 + 4^2 + 4^3 + ... + 4^2021 chia hết cho 21.
Giải thích
Ta có: \(A = 1 + 4 + {4^2} + {4^3} + ... + {4^{2021}}\)
\[ = \left( {1 + 4 + {4^2}} \right)\left( {{4^3} + {4^4} + {4^5}} \right) + ... + \left( {{4^{2019}} + {4^{2020}} + {4^{2021}}} \right)\]
\[ = \left( {1 + 4 + {4^2}} \right) + {4^3}\left( {1 + 4 + {4^2}} \right) + ... + {4^{2019}}\left( {1 + 4 + {4^2}} \right)\]
\[ = 21.(\left( {1 + {4^3} + ... + {4^{2019}}} \right)\]
Vì \(21\,\, \vdots \,\,21\) nên \(A\, \vdots \,\,21\)