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Maths question

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  • 31-05-2010 12:53pm
    #1
    Closed Accounts Posts: 55 ✭✭


    i got this in my mocks: prove by induction that 3 is a factor of 4(to the power of n) - 1. can anyone help me??


Comments

  • Registered Users Posts: 10,992 ✭✭✭✭partyatmygaff


    I can't help you as I haven't done any of that yet but if you want a better way of typing out maths on boards.ie just use the latex tags.

    [LATEX]Prove[/LATEX][LATEX]by[/LATEX][LATEX]induction[/LATEX] [LATEX]that[/LATEX] [LATEX]3[/LATEX] [LATEX]is[/LATEX] [LATEX] a[/LATEX] [LATEX]factor[/LATEX] [LATEX]of[/LATEX] [LATEX]4^n^-1[/LATEX]



    The tag is LATEX /LATEX with square brackets around both.


  • Registered Users Posts: 47 mark.oc


    RedRebel92 wrote: »
    i got this in my mocks: prove by induction that 3 is a factor of 4(to the power of n) - 1. can anyone help me??

    Skipping the prove true for n=1 section:

    Assume true for n=k:
    4^k - 1 = 3A
    4^k = 3A + 1

    Prove true for n=k+1:
    4^(k+1) - 1
    (4^k x 4^1) - 1
    (3A + 1)x4 - 1 (from step 1)
    12A + 4 - 1
    12A + 3
    which is divisible by 3.


  • Registered Users Posts: 10,992 ✭✭✭✭partyatmygaff


    mark.oc wrote: »
    Skipping the prove true for n=1 section:

    Assume true for n=k:
    [LATEX]4^k - 1 = 3A[/LATEX]
    [LATEX]4^k = 3A + 1[/LATEX]

    Prove true for n=k+1:
    [LATEX]4^(k+1) - 1[/LATEX]
    [LATEX](4^k x 4^1) - 1[/LATEX]
    [LATEX](3A + 1)*4 - 1 [/LATEX](from step 1)
    [LATEX]12A + 4 - 1[/LATEX]
    [LATEX]12A + 3[/LATEX]
    which is divisible by 3.
    Added latex tags to make it easier to read. Hope you don't mind.


  • Closed Accounts Posts: 834 ✭✭✭Reillyman


    Testing:

    [LATEX] x^2+3x+15=7^(n-1) = 2^k [/LATEX]


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