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Thread: mathematical induction

  1. #1
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    mathematical induction

    Can you please help me with
    Prove that n^3-n is divisible by 3
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by wantanswers View Post
    Can you please help me with
    Prove that n^3-n is divisible by 3
    $\displaystyle n^3 - n = 3a$ for some $\displaystyle a$.

    First we see if it holds for $\displaystyle n=0$

    $\displaystyle 0^3 - 0 = 0$, therefore true for $\displaystyle n=0$

    Now we use what's called the induction step, we assume that it's true for $\displaystyle n=k$ where k is some natural nubmer.

    This gives us $\displaystyle k^3 - k = 3a$

    We then prove this for $\displaystyle n=k+1$, giving us

    $\displaystyle (k+1)^3 - k - 1$.

    Expanding the brackets gives $\displaystyle k^3 + 3k^2 + 3k + 1 - k - 1$

    $\displaystyle k^3 + 3k^2 + 2k$ which can be written as $\displaystyle k^3 + 3k^2 + 3k - k$

    $\displaystyle (k^3-k) + 3(k^2 + k) = 3a + 3(k^2 + k)$.

    Therefore it is true for $\displaystyle k+1$ if true for $\displaystyle k$.

    You know it is true for 0, therefore true for 1. True for one, therefore true for 2....

    Hope this helps
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