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I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.

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  • $\begingroup$ I believe the upper bound is supposed to be $x-1$, but yes, that is correct. I'll explain in more detail in the answer. $\endgroup$
    – 2012ssohn
    Mar 3, 2014 at 13:42
  • $\begingroup$ Consider the "consecutive difference" $(x+1)^2-x^2$. What is the result? What about for $x^2-(x-1)^2$? Note that in general this is referred to as a "finite difference" on Wikipedia, etc. $\endgroup$
    – abiessu
    Mar 3, 2014 at 13:42
  • $\begingroup$ We know that $\sum_{k=0}^x k=\frac{k(k+1)}{2}$. $\endgroup$ Mar 3, 2014 at 13:43
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    $\begingroup$ I like the title (intended pun or not): "I found this odd relationship..." $\endgroup$
    – amWhy
    Mar 3, 2014 at 14:08

8 Answers 8

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How do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x+1$ should equal $x^2$

Hope this picture will provide you with the visual aid you need. :-)

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    $\begingroup$ In terms of understanding "why", this is probably the best descriptor. $\endgroup$
    – Simon Rose
    Mar 3, 2014 at 14:03
  • $\begingroup$ Most satisfying solution. Rigourous summations are boring. $\endgroup$
    – Guy
    Mar 3, 2014 at 14:25
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    $\begingroup$ This really helped me get the intuition. Thanks! $\endgroup$
    – undo_all
    Mar 3, 2014 at 16:36
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Yet another picture for illustration:

enter image description here

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Recall that:

$$\sum_{k=0}^{x}k = \frac{x(x+1)}{2}$$

Then

$$\sum_{k=0}^x(2k + 1) = 2\sum_{k=0}^x k + \sum_{k=0}^x1 = x(x+1) + (x+1) = x^2 + 2x + 1 \neq x^2$$

Instead, since $x^2 + 2x + 1= (x+1)^2$, then

$$\sum_{k=0}^x(2k + 1) = (x+1)^2$$

Using $x-1$ in place of $x$, then you have:

$$\sum_{k=0}^{x-1}(2k + 1) = x^2$$

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  • $\begingroup$ I like this one, seems nicely rigorous. Thanks for the help! $\endgroup$
    – undo_all
    Mar 3, 2014 at 16:38
  • $\begingroup$ Thank you very much again. $\endgroup$
    – Sebastiano
    Sep 28, 2020 at 23:27
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We prove this via induction.

Base case ($x = 1$): $$1^2 = \sum_{k=0}^{1-1} (2k+1) = \sum_{k=0}^0 (2k+1) = 2\cdot 0+1 = 1$$

Inductive step: Suppose it is true for some $x$. Now, we note that $$(x+1)^2 = x^2 + 2x + 1$$

and that

$$\sum_{k=0}^{x+1-1} (2k+1) = \sum_{k=0}^{x-1} (2k+1) + 2x+1$$

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    $\begingroup$ induction is not really intuition as asked for in the original question. $\endgroup$
    – Guy
    Mar 3, 2014 at 13:48
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    $\begingroup$ @Sabyasachi - the question asked for a proof, which I have provided. $\endgroup$
    – 2012ssohn
    Mar 3, 2014 at 13:52
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The standard proof without words is as follows:

1   12    123    1234    ...
    22    223    2234
          333    3334
                 4444
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  • $\begingroup$ what is this even supposed to mean? Not ridiculing, but you can consider adding an explanation. Doesn't look "standard" to me $\endgroup$
    – Guy
    Mar 3, 2014 at 13:49
  • $\begingroup$ @Sabyasachi: Calculate the area of the squares in two ways: side length squared is the same as what you get by going layer by layer. The number of squares in the last layer increases by $2$ at each step and is initially $1$. $\endgroup$
    – J. J.
    Mar 3, 2014 at 13:59
  • $\begingroup$ Editing that comment into your question will be good i think $\endgroup$
    – Guy
    Mar 3, 2014 at 14:07
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    $\begingroup$ @Sabyasachi: It's not traditional with proofs without words. Besides, Lucian posted the same proof with a nicer picture. $\endgroup$
    – J. J.
    Mar 3, 2014 at 14:14
  • $\begingroup$ I was just saying that your proof is brilliant and intuitively the best, but only once you make it clear what you mean by those numbers. It wasn't immediately obvious that you were drawing individual cells of a sqaure. Anyway +1 major upvote. $\endgroup$
    – Guy
    Mar 3, 2014 at 14:20
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Notice : $$\begin{align}(x + 1)^2 - x^2 &= x^2 + 2x + 1 - x^2 \\&= 2x + 1\end{align}$$

We take a summation on both sides and see that a lot of cancellation occurs on the LHS:

$$\sum_{k = 0}^{x-1}\left((x+1)^2 - x^2\right) = \sum_{k = 0}^{x-1}(2x+1)\\ (x -1 + 1)^2 - 0^2 = \sum_{k = 0}^{x-1}(2x+1)\\ x^2 = \sum_{k = 0}^{x-1}(2x+1)$$

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$\sum_{k=0}^{x-1}2k=\left(0+\left(x-1\right)\right)+\left(1+\left(x-2\right)\right)+\cdots+\left(\left(x-1\right)+0\right)=x\left(x-1\right)=x^{2}-x$

hence:

$\sum_{k=0}^{x-1}(2k+1)=\sum_{k=0}^{x-1}2k+x=x^2$

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  • $\begingroup$ @MarkBennet Thank you. $\endgroup$
    – drhab
    Mar 3, 2014 at 14:31
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I noticed this a few days ago and verified the first 1,000,000,000 with a program. If you FOIL it, it makes sense. For example:

(4+1)(4+1) = 4x4 + 4x1 + 4x1 + 1 = 25 

Obviously this is equal to 5x5. Notice that the last three terms are odd when added because

2x + 1

is consecutive odd numbers, and the first term is the previous square. If we generalize:

(X+1)(X+1) = X*X + X*1 + X*1 + 1 = X^2 + 2X + 1

Does this help?

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