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Math Help - Induction help

  1. #1
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    Induction help

    Use the mathematical induction to show the following. For all integer n=>2
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  2. #2
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  3. #3
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    Quote Originally Posted by slipz418 View Post
    Use the mathematical induction to show the following. For all integer n=>2
    Google is your friend.
    This exercise occurs so frequently that you can find it practically anywhere.
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  4. #4
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    \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}

    If this is so, then

    \sum_{k=1}^{n+1}k^2\ should\ =\frac{(n+1)([n+1]+1)(2[n+1]+1)}{6}

    therefore

    P(n)

    \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}

    P(n+1)

    \sum_{k=1}^{n+1}k^2=\frac{(n+1)(n+2)(2n+3)}{6}

    Proof

    Express P(n+1) in terms of P(n)
    so that if P(n) really is true, then P(n+1) must also be true
    (P(n) being true causes P(n+1) to be true).

    \sum_{k=1}^{n+1}k^2=\sum_{k=1}^nk^2+(n+1)^2

    which, if P(n) is true will equal

    \frac{n(n+1)(2n+1)}{6}+(n+1)(n+1)=\frac{n(n+1)(2n+  1)+6(n+1)(n+1)}{6}

    =\frac{(n+1)[n(2n+1)+6n+6]}{6}=\frac{(n+1)(2n^2+7n+6)}{6}

    =\frac{(n+1)(2n+3)(n+2)}{6}

    Therefore, if the sum is valid for n=1, it's valid for n=2,
    if it's valid for n=2, it's valid for n=3,
    if it's valid for n=3, then it's valid for n=4
    all the way to infinity,

    hence if it's valid for n=1, it's valid for all n.

    \frac{1(1+1)(2+1)}{6}=\frac{6}{6}=1=1^2

    Therefore the equation for the sum of squares is valid
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