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

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

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