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

Toán Lớp 6: Cho A = 2 + 2^2 + 2^3 + 2^4 + … + 2^98. Chứng minh rằng a) A chia hết cho 2; b) A chia hết cho 3; c) A chia cho 7 dư 6.

Tháng Mười 24, 2022 Toán học 2 Comments 0 views

Toán Lớp 6: Cho A = 2 + 2^2 + 2^3 + 2^4 + … + 2^98. Chứng minh rằng
a) A chia hết cho 2; b) A chia hết cho 3; c) A chia cho 7 dư 6.

Comments ( 2 )

  1. diemmyhoa
    diemmyhoa
    24 Tháng Mười, 2022 at 4:12 chiều
    Reply
    A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    a) Ta có:
    A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    A = 2 . (1 + 2 + $2^{2}$ + $2^{3}$ + … + $2^{97}$)
    Vậy A $\vdots$ 2
    b) Ta có:
    A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    A = (2 + $2^{2}$) + ($2^{3}$ + $2^{4}$) + … + ($2^{97}$ + $2^{98}$)
    A = (2 + $2^{2}$) + $2^{2}$ . (2 + $2^{2}$) + $2^{4}$ . (2 + $2^{2}$) + … + $2^{96}$ . (2 + $2^{2}$)
    A = 6 + $2^{2}$ . 6 + $2^{4}$ . 6 + … + $2^{96}$ . 6
    A = 6 . (1 + $2^{2}$ + $2^{4}$ + $2^{6}$ + … + $2^{96}$)
    Vậy A $\vdots$ 3
    c) Ta có:
    A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    A = (2 + $2^{2}$) + ($2^{3}$ + $2^{4}$ + $2^{5}$) + … + ($2^{96}$ + $2^{97}$ + $2^{98}$)
    A = 6 + $2^{2}$ . (2 + $2^{2}$ + $2^{3}$) + $2^{5}$ . (2 + $2^{2}$ + $2^{3}$) + … + $2^{95}$ . (2 + $2^{2}$ + $2^{3}$)
    A = 6 + $2^{2}$ . 14 + $2^{5}$ . 14 + … + $2^{95}$ . 14
    A = 6 + 14 . ($2^{2}$ + $2^{5}$ + $2^{8}$ + … + $2^{95}$)
    A = 6 + 2 . 7 . ($2^{2}$ + $2^{5}$ + $2^{8}$ + … + $2^{95}$)
    Vì 2 . 7 . ($2^{2}$ + $2^{5}$ + $2^{8}$ + … + $2^{95}$) $\vdots$ 7 nên 6 + 2 . 7 . ($2^{2}$ + $2^{5}$ + $2^{8}$ + … + $2^{95}$) chia 7 dư 6
    Vậy A chia 7 dư 6

  2. buithaianh98
    buithaianh98
    24 Tháng Mười, 2022 at 4:11 chiều
    Reply
    A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    a) A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    A = 2 . (1 + 2 + $2^{2}$ + $2^{3}$ + … + $2^{97}$)
    ⇒ A $\vdots$ 2
    b) A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    A = (2 + $2^{2}$) + ($2^{3}$ + $2^{4}$) + … + ($2^{97}$ + $2^{98}$)
    A = (2 + $2^{2}$) + $2^{2}$ . (2 + $2^{2}$) + $2^{4}$ . (2 + $2^{2}$) + … + $2^{96}$ . (2 + $2^{2}$)
    A = 6 + $2^{2}$ . 6 + $2^{4}$ . 6 + … + $2^{96}$ . 6
    A = 6 . (1 + $2^{2}$ + $2^{4}$ + $2^{6}$ + … + $2^{96}$)
    ⇒ A $\vdots$ 3
    c) A = 2 + $2^{2}$ + $2^{3}$ + … + $2^{98}$
    A = (2 + $2^{2}$) + ($2^{3}$ + $2^{4}$ + $2^{5}$) … ($2^{96}$ + $2^{97}$ + $2^{98}$)
    A = 6 + $2^{2}$ . ($2^{3}$ + $2^{4}$ + $2^{5}$) + $2^{3}$ . ($2^{3}$ + $2^{4}$ + $2^{5}$) + … + $2^{93}$ . ($2^{3}$ + $2^{4}$ + $2^{5}$)
    A = 6 + 56 + $2^{3}$ . 56 + … + $2^{93}$ . 56
    A = 6 + 56 . (1 + $2^{3}$ + $2^{6}$ + … + $2^{93}$)
    A = 6 + 8 . 7 . (1 + $2^{3}$ + $2^{6}$ + … + $2^{93}$)
    Do 8 . 7 . ($2^{3}$ + $2^{6}$ + $2^{9}$ + … + $2^{93}$) $\vdots$ 7 nên A chia 7 dư 6

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

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