13

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 $x^2$.

Jose M Serra
  • 2,583
undo_all
  • 255
  • 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 03 '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 03 '14 at 13:42
  • We know that $\sum_{k=0}^x k=\frac{k(k+1)}{2}$. – Pol van Hoften Mar 03 '14 at 13:43
  • 17
    I like the title (intended pun or not): "I found this odd relationship..." – amWhy Mar 03 '14 at 14:08

8 Answers8

61

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. :-)

Lucian
  • 48,334
  • 2
  • 83
  • 154
15

Yet another picture for illustration:

enter image description here

ccorn
  • 9,803
11

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:

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

the_candyman
  • 14,064
  • 4
  • 35
  • 62
6

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

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

2012ssohn
  • 3,827
5

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:

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

Simply Beautiful Art
  • 74,685
Yiyuan Lee
  • 14,435
5

The standard proof without words is as follows:

1   12    123    1234    ...
    22    223    2234
          333    3334
                 4444
J. J.
  • 9,432
  • what is this even supposed to mean? Not ridiculing, but you can consider adding an explanation. Doesn't look "standard" to me – Guy Mar 03 '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 03 '14 at 13:59
  • Editing that comment into your question will be good i think – Guy Mar 03 '14 at 14:07
  • 3
    @Sabyasachi: It's not traditional with proofs without words. Besides, Lucian posted the same proof with a nicer picture. – J. J. Mar 03 '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. – Guy Mar 03 '14 at 14:20
3

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

hence:

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

drhab
  • 151,093
1

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?

andrew
  • 11