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Math Help - simple proof. LHS = RHS

  1. #1
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    simple proof. LHS = RHS

    If i have \frac{1}{n} \sum_{k=1}^n k^2= \frac{1}{6}(n+1)(2n+1)

    where k=1,2,3....n

    how do i prove the LHS = RHS?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sirellwood View Post
    If i have \frac{1}{n} \sum_{k=1}^n k^2= \frac{1}{6}(n+1)(2n+1)

    where k=1,2,3....n

    how do i prove the LHS = RHS?
    I claim that \sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}.

    How do I prove this? Induction.

    1) It's clearly true for n=1.

    2) Assume it's true for n.

    3) Prove it's true for n+1:

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

    Spoiler:
    \frac{n(n+1)(2n+1)}{6}+(n+1)^2=\frac{2n^3+3n^2+n}{  6}+\frac{6(n^2+2n+1)}{6}=\frac{2n^3+9n^2+13n+6}{6} =\frac{(n+1)(n+1)(2n+3)}{6}=\frac{(n+1)((n+1)+1)(2  (n+1)+1)}{6}=\sum_{k=1}^{n+1} k^2

    which is the desired result.

    So \sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6} and therefore \frac{1}{n}\sum_{k=1}^n k^2=\frac{(n+1)(2n+1)}{6}
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