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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: A=2+2^2+2^3+…+2^2018 chứng minh A chia hết cho 6

Toán Lớp 6: A=2+2^2+2^3+…+2^2018 chứng minh A chia hết cho 6

Comments ( 2 )

  1. \text{Lời giải và giải thích chi tiết:}
    \text{6=3.2}
    \text{+)Chia hết cho 3}
    \text{A=2+$2^{2}$ +$2^{3}$ +…+$2^{2018}$  }
    \text{A=(2+$2^{2}$)+($2^{3}$+$2^{4}$)+…+($2^{2017}$+$2^{2018}$)  }
    \text{A=2(1+2)+$2^{3}$(1+2)+…+$2^{2017}$(1+2)  }
    \text{A=2.3+$2^{3}$.3+…+$2^{2017}$.3  }
    \text{A=3(2+$2^{3}$+…+$2^{2017}$) } 
    \text{⇒A\vdots3 (1) }
    \text{+)Chia hết cho 2 }
    \text{A=2+$2^{2}$ +$2^{3}$ +…+$2^{2018}$  }
    \text{2A=2(2+$2^{2}$ +$2^{3}$ +…+$2^{2018}$) }
    \text{2A=$2^{3}$+$2^{4}$+…+$2^{2019}$ }
    \text{2A-A=($2^{3}$+$2^{4}$+…+$2^{2019}$)-(2+$2^{2}$ +$2^{3}$ +…+$2^{2018}$)  }
    \text{A=$2^{2019}$-2 \vdots2 (2)  }
    \text{Từ (1);(2) ta suy ra A\vdots6.  }
    \text{Vậy A\vdots6  }

  2. Answer
    A = 2 + 2^{2} + 2^{3} + … + 2^{2018}
    A = (2 + 2^{2}) + (2^{3} + 2^{4}) + … + (2^{2017} + 2^{2018})
    A = 2 . (1 + 2) + 2^{3} . (1 + 2) + … + 2^{2017} . (1 + 2)
    A = 2 . 3 + 2^{3} . 3 + … + 2^{2017} . 3
    A = 3 . (2 + 2^{3} + … + 2^{2017}) \vdots 3 (1)
    _______________________
    A = 2 + 2^{2} + 2^{3} + … + 2^{2018}
    2A = 2 . (2 + 2^{2} + 2^{3} + … + 2^{2018})
    2A = 2^{2} + 2^{3} + 2^{4} + … + 2^{2019}
    2A – A = (2^{2} + 2^{3} + 2^{4} + … + 2^{2019}) – (2 + 2^{2} + 2^{3} + … + 2^{2018})
    A = 2^{2019} – 2 \vdots 2 (2)
    $\text{Từ}$ (1) ; (2) => A  \vdots 6
    $\text{Vậy bài toán được chứng minh}$ 

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222-9+11+12:2*14+14 = ? ( )