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

Toán Lớp 6: Chứng minh rằng: 2+2^2+2^3+2^4+…+2^60 chia hết cho 3;7;15

Toán Lớp 6: Chứng minh rằng:
2+2^2+2^3+2^4+…+2^60 chia hết cho 3;7;15

Comments ( 2 )

  1. A= (2+22)+(23+24)+…+(259+260)
    A=2.(1+2)+23.(1+2)+…+259.(1+2)
    A=2.3+23.3+…+259.3
    A=3.(2+23+…+259)
    Vì 3 chia hết cho 3 => 3.(2+23+…+259) chia hết cho 3
    =>A chia hết cho 3
    A= (2+22+23)+…+(258+259+260)
    A=2.(1+2+22)+…+258.(1+2+22)
    A=2.7+…+258.7
    A=7.(2+…+258)
    Vì 7 chia hết cho 7 =>7.(2+…+258) chia hết cho 7
    =>A chia hết cho 7
    A= (2+22+23+24)+…+(257+258+259+260)
    A=2.(1+2+22+23)+…+257.(1+2+22+23)
    A=2.15 +…+257.15
    A=15.(2+…+257)
    vì 15 chia hết cho15=>15.(2+…+25) chia hết cho 15
    =>A chia hết cho 15

  2. Chứng minh rằng: 2+2^2+2^3+2^4+…+2^60vdots3;7;15
    A= (2+2^2)+(2^3+2^4)+…+(2^59+2^60)
    A=2.(1+2)+2^3.(1+2)+…+2^59.(1+2)
    A=2.3+2^3. 3+…+2^59. 3
    A=3.(2+2^3+…+2^59)
    Vì 3vdots3 => 3.(2+2^3+…+2^59)vdots3
    =>Avdots3
    A= (2+2^2+2^3)+…+(2^58+2^59+2^60)
    A=2.(1+2+2^2)+…+2^58.(1+2+2^2)
    A=2.7+…+2^58. 7
    A=7.(2+…+2^58)
    Vì 7vdots7 =>7.(2+…+2^58)vdots7
    =>Avdots7
    A= (2+2^2+2^3+2^4)+…+(2^57+2^58+2^59+2^60)
    A=2.(1+2+2^2+2^3)+…+2^57.(1+2+2^2+2^3)
    A=2. 15 +…+2^57. 15
    A=15.(2+…+2^57)
    Vì 15vdots15=>15.(2+…+2^58)vdots15
    =>Avdots15

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