cho A= 2+2mu2+2mu3+2mu4+……+2mu60 de chia het cho 3 va5 va7

0 bình luận về “cho A= 2+2mu2+2mu3+2mu4+……+2mu60 de chia het cho 3 va5 va7”

1. Ta có :
   $\begin{array}{l}
   + )\,\,A = 2 + {2^2} + {2^3} + {2^4} + {….2^{60}}\\
   A = \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right) + … + \left( {{2^{59}} + {2^{60}}} \right)\\
   A = 2\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + … + {2^{59}}\left( {1 + 2} \right)\\
   A = 3.\left( {2 + {2^3} + … + {2^{59}}} \right)\,\, \vdots \,\,3\\
   \Rightarrow A\,\, \vdots \,\,3\\
   + )A = 2 + {2^2} + {2^3} + {2^4} + {….2^{60}}\\
   A = \left( {2 + {2^2} + {2^3}} \right) + \left( {{2^4} + {2^5} + {2^6}} \right) + … + \left( {{2^{58}} + {2^{59}} + {2^{60}}} \right)\\
   A = 2\left( {1 + 2 + {2^2}} \right) + {2^4}\left( {1 + 2 + {2^2}} \right) + … + {2^{58}}\left( {1 + 2 + {2^2}} \right)\\
   A = 2.7 + {2^4}.7 + … + {2^{58}}.7\\
   A = 7.\left( {2 + {2^4} + … + {2^{58}}} \right)\,\, \vdots \,\,7\\
   \Rightarrow A\,\, \vdots \,\,7
   \end{array}$

   Trả lời

2. Đáp án:

   A= 2+2^2+2^3+2^4+……+2^60

   = (2+2^2)+(2^3+2^4)+…+(2^59+2^60)

   = 2(1+2) + 2^3(1+2)+ …. + 2^59(1+2)

   = 2 . 3 + 2^3 . 3 +….+ 2^59 . 3

   = 3 ( 2+2^3+ … + 2^59) chia hết cho 3

   => A= 2+2^2+2^3+2^4+……+2^60 chia hết cho 3

   A= 2+2^2+2^3+2^4+……+2^60

   = (2+2^2+2^3)+ (2^4+2^5+2^6)+…+(2^58+2^59+2^60)

   = 2(1+2+2^2) + 2^4(1+2+2^2)+ …+ 2^58(1+2+2^2)

   = 2 . 7 + 2^4 . 7 +…+ 2^58 . 7

   = 7 ( 2+2^4+ …+ 2^58) chia hết cho 7

   => A= 2+2^2+2^3+2^4+……+2^60 chia hết cho 7

   A= 2+2^2+2^3+2^4+……+2^60

   = ( 2+2^2+2^3+2^4)+….+ ( 2^57+2^58+2^59+2^60)

   = 2(1+2+2^2+2^3)+ ….+ 2^57(1+2+2^2+2^3)

   = 2 . 15 + ….+ 2^57 . 15

   = 15(2+…+ 2^57) chia hết cho 5

   => A= 2+2^2+2^3+2^4+……+2^60 chia hết cho 5

   Trả lời