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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    5^{2n}-3^{n} is a multiple of 11 for all integers n\geq 1

    My attempt at a solution so far (I haven't included shown the case for P(1) etc.

    P(k):5^{2k}-3^{k}=11p \\ P(k+1):5^{2k+2}-3^{k+1} \\ 5^{2}\cdot 5^{2k}-3\cdot 3^{k} \\5^{2}( 3^{k}+11p)-3\cdot3^{k}

    I am not sure if it's correct up to here, and what the next step is.
    I have done the steps for P(1) to verify when n=1 (just not shown).
    Thank you for any help.

    P.S, I'm not used to using LaTeX
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  2. #2
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    Re: Mathematical Induction

    Quote Originally Posted by AW63 View Post
    5^{2n}-3^{n} is a multiple of 11 for all integers n\geq 1

    My attempt at a solution so far (I haven't included shown the case for P(1) etc.

    P(k):5^{2k}-3^{k}=11p \\ P(k+1):5^{2k+2}-3^{k+1}\\5^{2}\cdot 5^{2k}-3\cdot 3^{k}
    At this point use the common step of adding and subtracting the same thing to "separate" the parts:
    5^2\cdot 5^{2k}- 5^2\cdot 3^k+ 5^2\cdot 3^k- 3\cdot 3^k
    = 5^2(5^{2k}- 3^k)+ 3^k(25- 3)= 5^2(5^{2k}- 3^k})- 22(3^k)

     \\5^{2}( 3^{k}+11p)-3\cdot3^{k}

    I am not sure if it's correct up to here, and what the next step is.
    I have done the steps for P(1) to verify when n=1 (just not shown).
    Thank you for any help.

    P.S, I'm not used to using LaTeX
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  3. #3
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    Re: Mathematical Induction

    Quote Originally Posted by AW63 View Post
    5^{2n}-3^{n} is a multiple of 11 for all integers n\geq 1

    Assume we know that P(k):5^{2k}-3^{k}=11p is true.

    \begin{align*} 5^{2k+2}-3^{k+1}&= 5^{2k+2}-25\cdot 3^k + 25\cdot 3^{k}-3^{k+1}\\  &= 25(5^{2k}-3^{k})+3^k(25-3)   \end{align*}

    Can you finish?
    Last edited by Plato; December 22nd 2012 at 04:20 PM.
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  4. #4
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    Re: Mathematical Induction

    Thank you very much for your help, much appreciated.
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