12
$\begingroup$

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 $x^2$.

$\endgroup$
  • $\begingroup$ 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 '14 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 '14 at 13:42
  • $\begingroup$ We know that $\sum_{k=0}^x k=\frac{k(k+1)}{2}$. $\endgroup$ – user45878 Mar 3 '14 at 13:43
  • 16
    $\begingroup$ I like the title (intended pun or not): "I found this odd relationship..." $\endgroup$ – amWhy Mar 3 '14 at 14:08
58
$\begingroup$

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. :-)

| cite | improve this answer | |
$\endgroup$
  • 6
    $\begingroup$ In terms of understanding "why", this is probably the best descriptor. $\endgroup$ – Simon Rose Mar 3 '14 at 14:03
  • $\begingroup$ Most satisfying solution. Rigourous summations are boring. $\endgroup$ – Guy Mar 3 '14 at 14:25
  • 1
    $\begingroup$ This really helped me get the intuition. Thanks! $\endgroup$ – undo_all Mar 3 '14 at 16:36
13
$\begingroup$

Yet another picture for illustration:

enter image description here

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

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:

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I like this one, seems nicely rigorous. Thanks for the help! $\endgroup$ – undo_all Mar 3 '14 at 16:38
6
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ induction is not really intuition as asked for in the original question. $\endgroup$ – Guy Mar 3 '14 at 13:48
  • 4
    $\begingroup$ @Sabyasachi - the question asked for a proof, which I have provided. $\endgroup$ – 2012ssohn Mar 3 '14 at 13:52
5
$\begingroup$

The standard proof without words is as follows:

1   12    123    1234    ...
    22    223    2234
          333    3334
                 4444
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '14 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 '14 at 13:59
  • $\begingroup$ Editing that comment into your question will be good i think $\endgroup$ – Guy Mar 3 '14 at 14:07
  • 3
    $\begingroup$ @Sabyasachi: It's not traditional with proofs without words. Besides, Lucian posted the same proof with a nicer picture. $\endgroup$ – J. J. Mar 3 '14 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 '14 at 14:20
5
$\begingroup$

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:

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

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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

hence:

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @MarkBennet Thank you. $\endgroup$ – drhab Mar 3 '14 at 14:31
1
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.