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Thread: Is this sum property right?

  1. #1
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    Is this sum property right?

    $\displaystyle \sum_{i=1}^{n} (n + i)^2 = (\sum_{i=1}^{n - 1} (n - 1 + i)^2) + (2n - 1)^2 + (2n)^2 - n^2$

    Proof:

    $\displaystyle \sum_{i=1}^{n} (n + i)^2 = (n + 1)^2 + (n + 2)^2 + \ldots + (2n - 3)^2 + (2n - 2)^2 + (2n - 1)^2 + (2n)^2 = n^2 + (n + 1)^2 + (n + 2)^2 + \ldots + (2n - 3)^2 + (2n - 2)^2 + (2n - 1)^2 + (2n)^2 - n^2$
    $\displaystyle \sum_{i=1}^{n - 1} (n - 1 + i)^2 = n^2 + (n + 1)^2 + \ldots + (2n - 3)^2 + (2n - 2)^2$

    So, $\displaystyle \sum_{i=1}^{n} (n + i)^2 = \sum_{i=1}^{n - 1} (n - 1 + i)^2 + (2n - 1)^2 + (2n)^2 - n^2$
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  2. #2
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    Re: Is this sum property right?

    The answer you gave appears correct. Expanded, the result can be written as:
    $\displaystyle \sum_{i=1}^{n} (n + i)^2 = (\sum_{i=1}^{n - 1} (n - 1 + i)^2) + (2n - 1)^2 + (2n)^2 - n^2 = (\sum_{i=1}^{n - 1} (n - 1 + i)^2) + 7n^2 - 4n + 1$

    I used algebra instead of expanding each summation and got the same answer. Your answer is slicker, so if your proof is sufficient for your purpose, then you can stick by it.

    My way
    $\displaystyle \sum_{i=1}^{n} (n + i)^2 = \sum_{i=1}^{n-1} (n + i)^2 + (2n)^2$ (Take out the last term)
    $\displaystyle = [\sum_{i=1}^{n-1} (n - 1 + i)^2 + 2n-1 + 2i] + (2n)^2$ (Expanded each square and compared)
    $\displaystyle = \sum_{i=1}^{n-1} (n - 1 + i)^2 + 2 \sum_{i=1}^{n-1} i + (n-1)(2n-1) + (2n)^2$ (Removing (2n-1) from the sum means multiplying it n-1 times)
    $\displaystyle = \sum_{i=1}^{n-1} (n - 1 + i)^2 + (n-1)n + (n-1)(2n-1) + (2n)^2$

    The part outside the first sum is equal to what you have, which is
    $\displaystyle 7n^2 - 4n + 1$
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  3. #3
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    Re: Is this sum property right?

    double post
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