A=1+4+4^2+…+4^11.cmr A chia het cho 21 ,105 ,4097 29/07/2021 Bởi Gabriella A=1+4+4^2+…+4^11.cmr A chia het cho 21 ,105 ,4097
\(\begin{array}{l}A = 1 + 4 + {4^2} + ….. + {4^{11}}\\ = \left( {1 + 4 + {4^2}} \right) + \left( {{4^3} + {4^4} + {4^5}} \right) + \left( {{4^6} + {4^7} + {4^8}} \right) + \left( {{4^9} + {4^{10}} + {4^{11}}} \right)\\ = \left( {1 + 4 + {4^2}} \right) + {4^3}\left( {1 + 4 + {4^2}} \right) + {4^6}\left( {1 + 4 + {4^2}} \right) + {4^9}\left( {1 + 4 + {4^2}} \right)\\ = 21 + {4^3}.21 + {4^6}.21 + {4^9}.21\\ = 21\left( {1 + {4^3} + {4^6} + {4^9}} \right)\end{array}\) Vì \(21\,\, \vdots \,\,21 \Rightarrow A\,\, \vdots \,\,21.\) \(\begin{array}{l}A = 21\left( {1 + {4^3} + {4^6} + {4^9}} \right)\\ = 21\left[ {\left( {1 + {4^3}} \right) + {4^6}\left( {1 + {4^3}} \right)} \right]\\ = 21\left( {1 + {4^3}} \right)\left( {1 + {4^6}} \right)\\ = 21.65.4097\\ = 21.5.13.4097\end{array}\) \( \Rightarrow \left\{ \begin{array}{l}A{\kern 1pt} {\kern 1pt} \, \vdots \,\,105\\A{\kern 1pt} {\kern 1pt} \, \vdots \,\,4097\end{array} \right..\) Bình luận
\(\begin{array}{l}A = 1 + 4 + {4^2} + ….. + {4^{11}}\\ = \left( {1 + 4 + {4^2}} \right) + \left( {{4^3} + {4^4} + {4^5}} \right) + \left( {{4^6} + {4^7} + {4^8}} \right) + \left( {{4^9} + {4^{10}} + {4^{11}}} \right)\\ = \left( {1 + 4 + {4^2}} \right) + {4^3}\left( {1 + 4 + {4^2}} \right) + {4^6}\left( {1 + 4 + {4^2}} \right) + {4^9}\left( {1 + 4 + {4^2}} \right)\\ = 21 + {4^3}.21 + {4^6}.21 + {4^9}.21\\ = 21\left( {1 + {4^3} + {4^6} + {4^9}} \right)\end{array}\)
Vì \(21\,\, \vdots \,\,21 \Rightarrow A\,\, \vdots \,\,21.\)
\(\begin{array}{l}A = 21\left( {1 + {4^3} + {4^6} + {4^9}} \right)\\ = 21\left[ {\left( {1 + {4^3}} \right) + {4^6}\left( {1 + {4^3}} \right)} \right]\\ = 21\left( {1 + {4^3}} \right)\left( {1 + {4^6}} \right)\\ = 21.65.4097\\ = 21.5.13.4097\end{array}\)
\( \Rightarrow \left\{ \begin{array}{l}A{\kern 1pt} {\kern 1pt} \, \vdots \,\,105\\A{\kern 1pt} {\kern 1pt} \, \vdots \,\,4097\end{array} \right..\)