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

Toán Lớp 6: cho M= 3+3^2+3^3+…+3^2020+3^2021 Chứng minh rằng: M chia hết cho 13.

Tháng Ba 24, 2022 Toán học 2 Comments 0 views

Toán Lớp 6: cho M= 3+3^2+3^3+…+3^2020+3^2021
Chứng minh rằng: M chia hết cho 13.

Comments ( 2 )

  1. thanhbinh2k5
    thanhbinh2k5
    24 Tháng Ba, 2022 at 1:07 sáng
    Reply
    M=3+3^2+3^3+….+3^{2020}+3^{2021}
    =(3+3^2+3^3)+(3^4+3^5+3^6)+….+(3^{2019}+3^{2020}+3^{2021})
    =(3+3^2+3^3)+3^3.(3+3^2+3^3)+….+3^{2018}.(3+3^2+3^3)
    =(3+3^2+3^3).(1+3^3+…..+3^{2018})
    =(3+9+27).(1+3+3^3+…+3^{2018})
    =39.(1+3+3^3+…+3^{2018})
    $\\$
    Có 39 \vdots 13
    =>39.(1+3+…+3^{2018}) \vdots 13
    =>A \vdots 13
     

  2. cobelolen
    cobelolen
    24 Tháng Ba, 2022 at 1:07 sáng
    Reply
    A = 1 + 3 + 3^2 + 3^3 + . . . + 3^2021  
    = ( 1 + 3 + 3^2 ) + ( 3^3 + 3^4 + 3^5 ) + . . . + ( 3 ^2019 + 3^2020 + 3^2021 )
    = ( 1 + 3 + 3^2 ) + 3^3 ( 1 + 3 + 3^2 ) + . . . + 3^2019 ( 1 + 3 + 3^2 )
    = ( 1 + 3 + 3^2 ) ( 1 + 3^3 + . . . + 3^2019 )
    = 13. ( 1 + 3^3 + . . . + 3^2019 )
    Vì  13 vdots 13  nên  13. ( 1 + 3^3 + . . . + 3^2019 ) vdots 13
    Vậy  A vdots 13

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

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