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    Given that n is a positive integer, greater than 2, use the method of mathematical induction to prove that 2n>2n

    I've shown that its true for the base step n=3.

    Assume true for n=k
    =>2k>2k
    =>2k+1>2(k+1)

    Not sure what to do next :confused:
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    (Original post by bobbricks)
    Given that n is a positive integer, greater than 2, use the method of mathematical induction to prove that 2n>2n

    I've shown that its true for the base step n=3.

    Assume true for n=k
    =>2k>2k
    =>2k+1>2(k+1)

    Not sure what to do next :confused:

    I do not understand your structure


    Assume true for n=k
    =>2k>2k
    =>2k+1>4k =2k+2k>2k + 6 > 2k +2 ....
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    (Original post by TeeEm)
    I do not understand your structure


    Assume true for n=k
    =>2k>2k
    =>2k+1>4k =2k+2k>2k + 6 > 2k +2 ....
    Where did the 4k come from?
    Assume true for n=k
    =>2k>2k

    Then for n=k+1
    =>2k+1>2k+2
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    (Original post by bobbricks)
    Where did the 4k come from?
    multiplied both sides by 2
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    (Original post by bobbricks)
    Then for n=k+1
    =>2k+1>2k+2
    This is what you are trying to prove. You cannot just state it. You need to use 2^k?2k to prove that 2^{k+1}>2(k+1).

    You have only assumed that it is true for n=k, not for n=k+1.
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    Assume true for n=k
    => 2k > 2k

    Then for n=k+1
    => 2k+1 = 2*(2k) > 2*2k [from above]

    =>
    2k+1 > 2k + 2k > 2k + 2 [as k > 2]

    And so on...

    Key part is highlighted in bold
 
 
 
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