bài 1 Cho: A=1+4+4 mũ 2+...+4 mũ 23 Chứng minh rằng A chia hết cho 21

$\begin{array}{l} A = 1 + 4 + {4^2} + ... + {4^{23}}\\ \Rightarrow 4A = 4 + {4^2} + ... + {4^{24}}\\ \Rightarrow 4A - A = \left( {4 + {4^2} + ... + {4^{24}}} \right) - \left( {1 + 4 + {4^2} + ... + {4^{23}}} \right)\\ \Leftrightarrow 3A = {4^{24}} - 1\\ \Leftrightarrow A = \frac{{{4^{24}} - 1}}{3} = \frac{{{{\left( {{4^3}} \right)}^8} - 1}}{3}\\ Dat\,\,{4^3} = a,\,\,\,ta\,\,co:\\ A = \frac{{{a^8} - 1}}{3} = \frac{{\left( {{a^4} - 1} \right)\left( {{a^4} + 1} \right)}}{3} = \frac{{\left( {{a^2} - 1} \right).\left( {{a^2} + 1} \right).\left( {{a^4} + 1} \right)}}{3}\\ = \frac{{\left( {a - 1} \right).\left( {a + 1} \right).\left( {{a^2} + 1} \right).\left( {{a^4} + 1} \right)}}{3}\\ = \frac{{\left( {{4^3} - 1} \right).\left( {{4^3} + 1} \right).\left( {{4^6} + 1} \right).\left( {{a^{12}} + 1} \right)}}{3}\\ = \frac{{63.\left( {{4^3} + 1} \right).\left( {{4^6} + 1} \right).\left( {{a^{12}} + 1} \right)}}{3} = 21.\left( {{4^3} + 1} \right).\left( {{4^6} + 1} \right).\left( {{a^{12}} + 1} \right)\\ \Rightarrow A\,\, \vdots \,\,21\,\,\,\,(dpcm) \end{array}$