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

Toán Lớp 6: Chứng minh : A= 2^1+2^2+2^3+2^4+…+2^2020 chia hết cho 3 và 7

Tháng Tám 29, 2022 Toán học 2 Comments 0 views

Toán Lớp 6: Chứng minh : A= 2^1+2^2+2^3+2^4+…+2^2020 chia hết cho 3 và 7

Comments ( 2 )

  1. behocgioitoan
    behocgioitoan
    29 Tháng Tám, 2022 at 2:17 chiều
    Reply
    A= 2^1+2^2+2^3+2^4+…+2^2020
    ->A=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+…+(2^2019+2^2020)
    ->A=2.(1+2)+2^3(1+2)+2^5(1+2)+…+2^2019(1+2)
    ->A=2.3+2^3 . 3+2^5 . 3+..+2^2019 . 3
    ->A=3.(2+2^3+2^5+…+2^2019)
    ->A\vdots3
    Vậy A\vdots3
    A= 2^1+2^2+2^3+2^4+…+2^2020
    ->A=(2^1+2^2+2^3)+(2^4+2^5+2^6)+…+(2^2018+2^1019+2^2020)
    ->A=2(1+2+2^2)+2^4(1+2+2^2)+…+2^2018(1+2+2^2)
    ->A=2(1+2+4)+2^4(1+2+4)+…+2^2018(1+2+4)
    ->A=2.7+2^4 . 7+…+2^2018 .7
    ->A=7.(2+2^4+…+2^2018)
    ->A\vdots7
    Vậy A\vdots7

  2. thanhthuythu
    thanhthuythu
    29 Tháng Tám, 2022 at 2:17 chiều
    Reply
    Ta có : A = 2 + 2^2 + 2^3 + … + 2^2010
    = ( 2 + 2^2 ) + ( 2^3 + 2^4 ) + … + ( 2^2009 + 2^2010 )
    = 2 ( 1 + 2 ) + 2^3 ( 1 + 2 ) + … + 2^2009 ( 1 + 2 )
    = 2 . 3 + 2^3 . 3 + … + 2^2009 . 3
    Vì 3⋮⋮3 nên 2 . 3 + 2^3 . 3 + … + 2^2009 . 3 ⋮⋮ 3
    hay A ⋮⋮ 3   (đpcm)
    Ta có : A = 2 + 2^2 + 2^3 + … + 2^2010
    = ( 2 + 2^2 + 2^3 ) + ( 2^4 + 2^5 + 2^6 ) + …  + ( 2^2008 + 2^2009 + 2^2010 )
    = 2 ( 1 + 2 + 2^2 ) + 2^4 ( 1 + 2 + 2^2 ) + … + 2^2008 ( 1 + 2 + 2^2 )
    = 2 . 7 + 2^4 . 7 + … + 2^2008 . 7
    Vì 7 ⋮⋮ 7 nên 2 . 7 + 2^4 . 7 + … + 2^2008 . 7 ⋮⋮ 7
    ​hay A ⋮⋮ 7  (đpcm)

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

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