S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 chứng minh biểu thức S chia hết cho 14

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S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 chứng minh biểu thức S chia hết cho 14

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Jasmine 5 tháng 2021-07-23T08:36:46+00:00 2 Answers 8 views 0

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    0
    2021-07-23T08:37:58+00:00

    S= 2+2²+2³+…+$2^{99}$

    = ( 2+2²+2³)+…+( $2^{97}$+$2^{98}$$2^{99}$)

    = 14+…+$2^{96}$.( 2+2²+2³)

    = 14+…+$2^{96}$.14

    = 14.( 1+…+$2^{96}$)⋮ 14

    Vậy S⋮ 14

    0
    2021-07-23T08:38:37+00:00

    S=2+2^2+2^3+2^4+2^5+…+2^98+2^99

      =(2+2^2+2^3)+(2^4+2^5+2^6+2^7)+…+(2^96+2^97+2^98+2^99)

      =(2+2^2+2^3)+2^4(2+2^2+2^3)+…+2^96(2+2^2+2^3)

      =(2+2^2+2^3)(1+2^4+….+2^96)

      =14.(1+2^4+….+2^96) ⋮ 14

    =>S ⋮ 14

    Vậy biểu thức S chia hết cho 14.

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