chon kit 發問於 科學及數學數學 · 1 十年前

123A-MATHS

prove,by MI 3^2n-2^2n is divisble by 5

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  • 1 十年前
    最愛解答

    3^(2(1)) - 2^(2(1)) = 5 is divisble by 5

    assume 3^(2k) - 2^(2k) is divisble by 5

    3^(2k) - 2^(2k) = 5m, where m is integer

    consider

    3^(2(k+1)) - 2^(2(k+1))

    = (9)3^(2k) - (4)2^(2k)

    = (9)[ 5m + 2^(2k) ] - (4)2^(2k)

    = (5) 9m + (5)2^(2k)

    = (5) (9m + 2^(2k) )

    because 9m + 2^(2k) is integer, 3^(2(k+1)) - 2^(2(k+1)) is divisble by 5

    so, when case n = k is true, case n = k+1 is also true

    2007-11-03 19:41:32 補充:

    慢左, 不過請投我一票, 最多下次又幫過你

  • NCY
    Lv 7
    1 十年前

    雞尾包拉票又再一次成功.........

  • 1 十年前

    以自我宣傳形式的最佳回答是不行的, 請接受公平競技

  • 1 十年前

    若能以公平的態度去對待兩位的回答則更好。

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  • 1 十年前

    Let P ( n ) be the proposition “32n-22n is divisble by 5”.

    When n = 1,

    32-22=5 which is divisible by 5, so P ( 1 ) is true.

    Assume P ( k ) is true for some positive integers k, i.e.

    32k-22k=5M where M is an integer

    When n = k + 1,

    32k+2-22k+2

    =9(32k)-4(22k)

    =9(32k-22k)-4(22k)+9(22k)

    =9(5M)+5(22k)

    =5(9M+22k)

    So P ( k + 1 ) is true.

    By MI, P ( n ) is true for all positive integers n.

    2007-11-05 18:44:00 補充:

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    2008-06-26 17:58:48 補充:

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    資料來源: My Maths Knowledge
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