# I found this odd relationship, $x^2 = \sum_{k = 0}^{x-1} (2k + 1)$.

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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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. – 2012ssohn Mar 3 '14 at 13:42
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. – abiessu Mar 3 '14 at 13:42
?Not true try with x=0. LHS = 0. RHS = 1. – oks Mar 3 '14 at 13:42
@oks - the problem has been fixed. – 2012ssohn Mar 3 '14 at 13:45
I like the title (intended pun or not): "I found this odd relationship..." – amWhy Mar 3 '14 at 14:08

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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In terms of understanding "why", this is probably the best descriptor. – Simon Rose Mar 3 '14 at 14:03
Most satisfying solution. Rigourous summations are boring. – Sabyasachi Mar 3 '14 at 14:25
This really helped me get the intuition. Thanks! – undo_all Mar 3 '14 at 16:36

Yet another picture for illustration:

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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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I like this one, seems nicely rigorous. Thanks for the help! – undo_all Mar 3 '14 at 16:38

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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induction is not really intuition as asked for in the original question. – Sabyasachi Mar 3 '14 at 13:48
@Sabyasachi - the question asked for a proof, which I have provided. – 2012ssohn Mar 3 '14 at 13:52

The standard proof without words is as follows:

1   12    123    1234    ...
22    223    2234
333    3334
4444

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what is this even supposed to mean? Not ridiculing, but you can consider adding an explanation. Doesn't look "standard" to me – Sabyasachi Mar 3 '14 at 13:49
@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 $1$. – J. J. Mar 3 '14 at 13:59
Editing that comment into your question will be good i think – Sabyasachi Mar 3 '14 at 14:07
@Sabyasachi: It's not traditional with proofs without words. Besides, Lucian posted the same proof with a nicer picture. – J. J. Mar 3 '14 at 14:14
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. – Sabyasachi Mar 3 '14 at 14:20

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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@MarkBennet Thank you. – drhab Mar 3 '14 at 14:31