Cho A = 2 +2 mũ 2 +2 mũ 3+ …..+2 mũ 60 Chứng minh A chia hết cho 3;7;15 19/07/2021 Bởi Piper Cho A = 2 +2 mũ 2 +2 mũ 3+ …..+2 mũ 60 Chứng minh A chia hết cho 3;7;15
Giải thích các bước giải: Ta có: \(\begin{array}{l}A = 2 + {2^2} + {2^3} + …. + {2^{60}}\\ = \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right) + \left( {{2^5} + {2^6}} \right) + ….. + \left( {{2^{59}} + {2^{60}}} \right)\\ = 2.\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + {2^5}\left( {1 + 2} \right) + …. + {2^{59}}\left( {1 + 2} \right)\\ = \left( {1 + 2} \right)\left( {2 + {2^3} + {2^5} + …. + {2^{59}}} \right)\\ = 3.\left( {2 + {2^3} + {2^5} + …. + {2^{59}}} \right) \vdots 3\\A = 2 + {2^2} + {2^3} + …. + {2^{60}}\\ = \left( {2 + {2^2} + {2^3}} \right) + \left( {{2^4} + {2^5} + {2^6}} \right) + ….. + \left( {{2^{58}} + {2^{59}} + {2^{60}}} \right)\\ = 2.\left( {1 + 2 + {2^2}} \right) + {2^4}\left( {1 + 2 + {2^2}} \right) + ….. + {2^{58}}\left( {1 + 2 + {2^4}} \right)\\ = \left( {1 + 2 + {2^2}} \right)\left( {2 + {2^4} + {2^7} + …. + {2^{58}}} \right)\\ = 7.\left( {2 + {2^4} + {2^7} + …. + {2^{58}}} \right) \vdots 7\\A = 2 + {2^2} + {2^3} + …. + {2^{60}}\\ = \left( {2 + {2^2} + {2^3} + {2^4}} \right) + \left( {{2^5} + {2^6} + {2^7} + {2^8}} \right) + …. + \left( {{2^{57}} + {2^{58}} + {2^{59}} + {2^{60}}} \right)\\ = 2\left( {1 + 2 + {2^2} + {2^3}} \right) + {2^5}\left( {1 + 2 + {2^2} + {2^3}} \right) + ….. + {2^{57}}\left( {1 + 2 + {2^2} + {2^3}} \right)\\ = \left( {1 + 2 + {2^2} + {2^3}} \right)\left( {2 + {2^5} + …. + {2^{57}}} \right)\\ = 15.\left( {2 + {2^5} + …. + {2^{57}}} \right) \vdots 15\end{array}\) Bình luận
$ A= 2+2²+2³+…+2^{60}$ ⇔ $ A= ( 2+2²)+…+( 2^{59}+2^{60})$ ⇔ $ A= 2.( 1+2)+…+2^{59}.( 1+2)$ ⇔ $ A= 2.3+…+2^{59}.3$ ⇔ $ A= 3.( 2+..+2^{59})$ ⇒ A⋮ 3 $ A= 2+2²+2³+…+2^{60}$ ⇔ $ A= ( 2+2²+2³)+…+( 2^{58}+2^{59}2^{60}) $ ⇔ $ A= 2.( 1+2+2²)+…+2^{58}.( 1+2+2²) $ ⇔ $ A= 2.7+…+2^{58}.7$ ⇔ $ A= 7.( 2+…+2^{58}$ ⇒ A⋮ 7 $ A= 2+2²+2³+…+2^{60}$ ⇔ $ A= ( 2+2²+2³+2^{4})+…+( 2^{57}+2^{58}+2^{59}+2^{60})$ ⇔ $ A= 2.( 1+2+2²+2³)+…+2^{57}.( 1+2+2²+2³)$ ⇔ $ A= 2.15+…+2^{57}.15$ ⇔ $ A= 15.( 2+…+2^{57}$ ⇒ A⋮ 15 Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = 2 + {2^2} + {2^3} + …. + {2^{60}}\\
= \left( {2 + {2^2}} \right) + \left( {{2^3} + {2^4}} \right) + \left( {{2^5} + {2^6}} \right) + ….. + \left( {{2^{59}} + {2^{60}}} \right)\\
= 2.\left( {1 + 2} \right) + {2^3}\left( {1 + 2} \right) + {2^5}\left( {1 + 2} \right) + …. + {2^{59}}\left( {1 + 2} \right)\\
= \left( {1 + 2} \right)\left( {2 + {2^3} + {2^5} + …. + {2^{59}}} \right)\\
= 3.\left( {2 + {2^3} + {2^5} + …. + {2^{59}}} \right) \vdots 3\\
A = 2 + {2^2} + {2^3} + …. + {2^{60}}\\
= \left( {2 + {2^2} + {2^3}} \right) + \left( {{2^4} + {2^5} + {2^6}} \right) + ….. + \left( {{2^{58}} + {2^{59}} + {2^{60}}} \right)\\
= 2.\left( {1 + 2 + {2^2}} \right) + {2^4}\left( {1 + 2 + {2^2}} \right) + ….. + {2^{58}}\left( {1 + 2 + {2^4}} \right)\\
= \left( {1 + 2 + {2^2}} \right)\left( {2 + {2^4} + {2^7} + …. + {2^{58}}} \right)\\
= 7.\left( {2 + {2^4} + {2^7} + …. + {2^{58}}} \right) \vdots 7\\
A = 2 + {2^2} + {2^3} + …. + {2^{60}}\\
= \left( {2 + {2^2} + {2^3} + {2^4}} \right) + \left( {{2^5} + {2^6} + {2^7} + {2^8}} \right) + …. + \left( {{2^{57}} + {2^{58}} + {2^{59}} + {2^{60}}} \right)\\
= 2\left( {1 + 2 + {2^2} + {2^3}} \right) + {2^5}\left( {1 + 2 + {2^2} + {2^3}} \right) + ….. + {2^{57}}\left( {1 + 2 + {2^2} + {2^3}} \right)\\
= \left( {1 + 2 + {2^2} + {2^3}} \right)\left( {2 + {2^5} + …. + {2^{57}}} \right)\\
= 15.\left( {2 + {2^5} + …. + {2^{57}}} \right) \vdots 15
\end{array}\)
$ A= 2+2²+2³+…+2^{60}$
⇔ $ A= ( 2+2²)+…+( 2^{59}+2^{60})$
⇔ $ A= 2.( 1+2)+…+2^{59}.( 1+2)$
⇔ $ A= 2.3+…+2^{59}.3$
⇔ $ A= 3.( 2+..+2^{59})$
⇒ A⋮ 3
$ A= 2+2²+2³+…+2^{60}$
⇔ $ A= ( 2+2²+2³)+…+( 2^{58}+2^{59}2^{60}) $
⇔ $ A= 2.( 1+2+2²)+…+2^{58}.( 1+2+2²) $
⇔ $ A= 2.7+…+2^{58}.7$
⇔ $ A= 7.( 2+…+2^{58}$
⇒ A⋮ 7
$ A= 2+2²+2³+…+2^{60}$
⇔ $ A= ( 2+2²+2³+2^{4})+…+( 2^{57}+2^{58}+2^{59}+2^{60})$
⇔ $ A= 2.( 1+2+2²+2³)+…+2^{57}.( 1+2+2²+2³)$
⇔ $ A= 2.15+…+2^{57}.15$
⇔ $ A= 15.( 2+…+2^{57}$
⇒ A⋮ 15