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Thread: induction

  1. #1
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    induction

    Hi,

    We're on induction right now. Can somebody help me?

    Prove that for all integers n, 3^(3n+1) + 2^(n+1) is a multiple of 5.
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  2. #2
    Super Member wingless's Avatar
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    $\displaystyle f(n) = 3^{3n+1}+2^{n+1}$

    $\displaystyle g(n)$ = Remainder of $\displaystyle \frac{f(n)}{5}$

    We'll prove that $\displaystyle g(n) = 0$ for all $\displaystyle n$ integers.

    Firstly, try it for an integer, for example $\displaystyle n=0$..
    $\displaystyle f(0) = 3^{1}+2^{1} = 5$
    $\displaystyle g(0) = 0$
    It satisfies.

    Now we can use induction.

    $\displaystyle f(n+1) = 3^{3n+4}+2^{n+2}$
    $\displaystyle f(n+1) = f(n) + 26\cdot 3^{3n+1} + 2^{n+1}$
    $\displaystyle f(n+1) = f(n) + \underbrace{25\cdot 3^{3n+1}}_{\text{Multiple of 5}} + \underbrace{3^{3n+1} + 2^{n+1}}_{\text{Multiple of 5}}$

    So if f(n) is divisible, f(n+1) is divisible too.

    Note: This doesn't work for all n integers. It works for $\displaystyle n\geq 0$
    Last edited by wingless; Mar 13th 2008 at 08:34 AM.
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