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Math Help - Mathematical Induction

  1. #1
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    Mathematical Induction

    I was wondering if someone can show me step by step on how to do this question:
    Prove that for all integers n >= 1, 6|n (n^2 + 5)
    So far I got the bases step for when n = 1.
    I am having troubles with the induction step after n = 1.
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  2. #2
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    Re: Mathematical Induction

    Quote Originally Posted by xboxwee View Post
    I was wondering if someone can show me step by step on how to do this question:
    Prove that for all integers n >= 1, 6|n (n^2 + 5)
    So far I got the bases step for when n = 1.
    I am having troubles with the induction step after n = 1.
    It's always more sophisticated to ask your questions in English instead of mathematical symbols.

    Just so we're clear, you're asking to prove that for all positive integers, that 6 divides n(n^2 + 5)?
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    Re: Mathematical Induction

    Yes.
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  4. #4
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    Re: Mathematical Induction

    The base step is obvious.

    For the inductive step, assume that 6 divides \displaystyle \begin{align*} k(k^2 + 5) \end{align*} , in other words, write \displaystyle \begin{align*}  k(k^2 + 5) = 6m \end{align*} where \displaystyle \begin{align*} m \end{align*} is some other positive integer.

    Then we need to show that 6 divides \displaystyle \begin{align*} (k + 1)\left[(k + 1)^2 + 5\right] \end{align*}

    \displaystyle \begin{align*} (k + 1)\left[(k + 1)^2 + 5\right] &= (k + 1)\left(k^2 + 2k + 1 + 5\right) \\ &= k\left(k^2 + 5 + 2k + 1\right) + 1\left(k^2 + 5 + 2k + 1\right) \\ &= k\left(k^2 + 5\right) + k(2k + 1) + k^2 + 2k + 6 \\ &= 6m + 2k^2 + k + k^2 + 2k + 6 \\ &= 6m + 3k^2 + 3k + 6 \\ &= 6m + 3k(k + 1) + 6 \\ &= 6m + 3\cdot 2p + 6 \textrm{ where }p \textrm{ is some other positive integer, which we can do since }k(k + 1) \textrm{ is even...} \\ &= 6m + 6p + 6 \\ &= 6\left(m + p + 1\right) \textrm{ which is clearly divisible by }6 \end{align*}

    Q.E.D.
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    Re: Mathematical Induction

    would this be right?
    Attached Thumbnails Attached Thumbnails Mathematical Induction-q3.jpg  
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    Re: Mathematical Induction

    Thanks..
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    Re: Mathematical Induction

    Quote Originally Posted by xboxwee View Post
    would this be right?
    No it's not a complete proof, as you have not shown that your final line is divisible by 6. In fact, you have not used your assumption at all.

    Also, the writing of your base step is atrocious.

    \displaystyle \begin{align*} 6|n\left(n^2 + 5\right) \end{align*} is a STATEMENT, that 6 divides \displaystyle \begin{align*} n\left(n^2 + 5\right) \end{align*} . It can NOT be used in equations.

    You would need to write something like

    If \displaystyle \begin{align*} n = 1 \end{align*} then \displaystyle \begin{align*} n\left(n^2 + 5\right) = 1\left(1^2 + 5\right) = 6 \end{align*} which is divisible by 6.
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