Toán Cho M =2+2^2+2^3+…+2^20. Chứng tỏ rằng M chia hết cho 5 06/09/2021 By Rylee Cho M =2+2^2+2^3+…+2^20. Chứng tỏ rằng M chia hết cho 5
Đáp án: Giải thích các bước giải: \(\begin{array}{l}M = 2 + {2^2} + {2^3} + {2^4} + …. + {2^{20}}\\M = \left( {2 + {2^2} + {2^3} + {2^4}} \right) + …. + \left( {{2^{17}} + {2^{18}} + {2^{19}} + {2^{20}}} \right)\\M = 2\left( {1 + 2 + {2^2} + {2^3}} \right) + … + {2^{17}}\left( {1 + 2 + {2^2} + {2^3}} \right)\\M = 2.15 + …. + {2^{17}}.15\\M = 15.\left( {2 + … + {2^{17}}} \right)\,\, \vdots \,5\end{array}\) Trả lời
\[\begin{array}{l}M = 2 + {2^2} + {2^3} + … + {2^{20}}\\M = \left( {{2^2} + {2^4}} \right) + \left( {{2^6} + {2^8}} \right) + … + \left( {{2^{18}} + {2^{20}}} \right) + \left( {2 + {2^3}} \right) + \left( {{2^5} + {2^7}} \right) + … + \left( {{2^{17}} + {2^{19}}} \right)\\M = {2^2}\left( {1 + {2^2}} \right) + {2^6}\left( {1 + {2^2}} \right) + … + {2^{18}}\left( {1 + {2^2}} \right) + 2\left( {1 + {2^2}} \right) + {2^5}\left( {1 + {2^2}} \right) + … + {2^{17}}\left( {1 + {2^2}} \right)\\M = {2^2}.5 + {2^6}.5 + … + {2^{18}}.5 + 2.5 + {2^5}.5 + … + {2^{17}}.5\\M = \left( {{2^2} + {2^6} + … + {2^{18}} + 2 + {2^5} + … + {2^{17}}} \right).5\\ \Rightarrow M\,\,\, \vdots \,\,5\end{array}\] Trả lời
Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}M = 2 + {2^2} + {2^3} + {2^4} + …. + {2^{20}}\\M = \left( {2 + {2^2} + {2^3} + {2^4}} \right) + …. + \left( {{2^{17}} + {2^{18}} + {2^{19}} + {2^{20}}} \right)\\M = 2\left( {1 + 2 + {2^2} + {2^3}} \right) + … + {2^{17}}\left( {1 + 2 + {2^2} + {2^3}} \right)\\M = 2.15 + …. + {2^{17}}.15\\M = 15.\left( {2 + … + {2^{17}}} \right)\,\, \vdots \,5\end{array}\)
\[\begin{array}{l}
M = 2 + {2^2} + {2^3} + … + {2^{20}}\\
M = \left( {{2^2} + {2^4}} \right) + \left( {{2^6} + {2^8}} \right) + … + \left( {{2^{18}} + {2^{20}}} \right) + \left( {2 + {2^3}} \right) + \left( {{2^5} + {2^7}} \right) + … + \left( {{2^{17}} + {2^{19}}} \right)\\
M = {2^2}\left( {1 + {2^2}} \right) + {2^6}\left( {1 + {2^2}} \right) + … + {2^{18}}\left( {1 + {2^2}} \right) + 2\left( {1 + {2^2}} \right) + {2^5}\left( {1 + {2^2}} \right) + … + {2^{17}}\left( {1 + {2^2}} \right)\\
M = {2^2}.5 + {2^6}.5 + … + {2^{18}}.5 + 2.5 + {2^5}.5 + … + {2^{17}}.5\\
M = \left( {{2^2} + {2^6} + … + {2^{18}} + 2 + {2^5} + … + {2^{17}}} \right).5\\
\Rightarrow M\,\,\, \vdots \,\,5
\end{array}\]