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Is this correct? If so a link to more information/proof would be appreciated. Thanks!

$$\sum_{k=1}^x(k + k - 1) = x^2$$

WolframAlpha

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    $\begingroup$ Remember what Sum_{k=1}^x k is. Use it and calculate, Mathematica not needed here. $\endgroup$ – mgamer Oct 28 '17 at 21:17
  • $\begingroup$ Awesome! Thanks mgamer for bringing it down to basics. x^2/2 + x/2 + x^2/2 + x/2 -x = x^2 $\endgroup$ – ID10T_ERROR Oct 28 '17 at 21:59
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You can do this one by induction.

The base case is that $\sum_{k=1}^1(2k - 1) = 1^2$ and this is pretty much self evident. Supposing it holds for the first $n$ positive integers means that:

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

Hence it holds for all positive integers.

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