13
$\begingroup$

The formula basically is:

The sum of all integers before and including $n$, plus all the integers up to and including $n-1$.

This will find $n^2$.

$$ \sum^n_{i = 1}i + \sum^{n-1}_{i=1}i = n^2 $$

$\endgroup$
4
  • 1
    $\begingroup$ You can write formulae on Math SE using TeX. This document should be enough to get you started: ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf $\endgroup$
    – 727
    Aug 5, 2015 at 22:45
  • $\begingroup$ From Wikipedia: "Most simply, the sum of two consecutive triangular numbers is a square number." See the images here. $\endgroup$ Aug 5, 2015 at 23:01
  • 1
    $\begingroup$ The sum of all positive integers. $\endgroup$
    – user26486
    Aug 7, 2015 at 4:33
  • $\begingroup$ In the title you ask whether somebody can check your proof. But there is no proof in your post and you have not poster a proof in an answer, either...? $\endgroup$ Aug 8, 2015 at 8:51

9 Answers 9

73
+50
$\begingroup$

$$\begin{array}{ccccccc}&&&\square&&&\\ &&\blacksquare&\square&\square\\ &\blacksquare&\blacksquare&\square&\square&\square\\ \blacksquare&\blacksquare&\blacksquare&\square&\square&\square&\square \end{array} \left.\rightarrow\quad \begin{array}{cccc} \square&\blacksquare&\blacksquare&\blacksquare\\ \square&\square&\blacksquare&\blacksquare\\ \square&\square&\square&\blacksquare\\ \square&\square&\square&\square \end{array}\quad\right\}n\\ $$


In numbers,

$$\underbrace{\begin{array}{lrrrrrrrrr} &n&+&n-1&+&n-2&+&\cdots&+&1\\ +&0&+&1&+&2&+&\cdots&+&n-1\\ \hline &n&+&n&+&n&+&\cdots&+&n \end{array}}_n$$


In summation signs,

$$\begin{align*} \sum_{i=1}^ni + \sum_{i=1}^{n-1}i &= \sum_{i=1}^ni + \sum_{i=0}^{n-1}i\\ &= \sum_{i=1}^ni + \sum_{j=1}^{n}(n-j) & (j = n-i)\\ &= \sum_{i=1}^n(i+n-i)\\ &= \sum_{i=1}^n n\\ &= n^2 \end{align*}$$

$\endgroup$
6
  • $\begingroup$ I just saw an arrow at first before realizing this is exactly the answer I had in mind. $\endgroup$ Aug 6, 2015 at 4:04
  • $\begingroup$ Maybe nicer: draw the little circles in a square. Then tilt your head diagonally and read off the diagonals. $\endgroup$ Aug 6, 2015 at 13:54
  • $\begingroup$ The colour in your first proof appears at first when I refresh the page, then disappears as the TeX renders. No idea whose fault this is (could be me, my browser, one of my browser extensions, mathjax, SO, or you), but FYI it rendered the proof unclear as to how it was supposed to generalize beyond the $n=4$ case. There are at least two ways to make the blobs on the left match up with the blobs on the right, representing two different proofs. Well, I suppose there are $16!$ if we ignore all symmetries, but 2 good ones I can immediately think of... $\endgroup$ Aug 6, 2015 at 17:16
  • 1
    $\begingroup$ FWIW, it renders properly over here. $\endgroup$
    – Brian Tung
    Aug 6, 2015 at 20:44
  • $\begingroup$ @SteveJessop Changed to b/w, see if it helps. $\endgroup$
    – peterwhy
    Aug 6, 2015 at 20:54
25
$\begingroup$

It is known that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$ Thus the value of your sum would be $$\sum_{k=1}^nk+\sum_{k=1}^{n-1}k=\frac{n(n+1)}{2}+\frac{(n-1)(n)}{2}=\frac{n^2+n+n^2-n}{2}=\frac{2n^2}{2}=n^2.$$

$\endgroup$
1
  • 2
    $\begingroup$ no intuition whatsoever but it's rigorous :) $\endgroup$ Aug 6, 2015 at 4:05
16
$\begingroup$

This is equivalent to the well-known fact that the sum of the first $n$ odd numbers is $n^2$. For example, $1+3+5+7+9+11=36$. Why are they equivalent? Because of this: \begin{align} 1+2+3+4+5+\phantom16&\\ {}+1+2+3+4+\phantom15&\\ -----------&\\ 1+3+5+7+9+11& \end{align}

$\endgroup$
3
  • 4
    $\begingroup$ See this image to see why the sum of the first $n$ odd numbers in $n^2$. (It's a proof-without-words.) $\endgroup$ Aug 5, 2015 at 22:57
  • $\begingroup$ I would say both sums are well-known. $\endgroup$
    – peterwhy
    Aug 5, 2015 at 22:59
  • $\begingroup$ neat way to think about it $\endgroup$ Aug 6, 2015 at 4:05
2
$\begingroup$

Note that $$\begin{align}i^2-(i-1)^2&\color{lightgray}{=2i-1}\\&=i\qquad+(i-1)\quad\quad\end{align}$$ Summing from $i=1$ to $n$ and telescoping LHS gives $$\begin{align}n^2\qquad\quad&=\sum_{i=1}^ni+\sum_{i=1}^n(i-1)\\ &=\sum_{i=1}^n i+\sum_{i=0}^{n-1}i\\ &=\sum_{i=1}^n i+\sum_{i=1}^{n-1}i\qquad\blacksquare\end{align}$$

$\endgroup$
1
$\begingroup$

Assuming you consider

$$ \sum^n_{i = 1}i = \frac{n(n+1)}{2} $$

to be a well-known fact, observe that your sum is just

$$ \begin{array}{rcl} \sum^n_{i = 1}i + \sum^{n-1}_{i=1}i & = & \sum^n_{i=1}i + \sum^n_{i=1}i - n \\ &=& 2\sum^n_{i=1}i - n\\ &=& 2\frac{n(n+1)}{2} - n\\ &=& n(n+1) - n \\ &=& n^2 + n - n \\ &=& n^2 \end{array} $$

$\endgroup$
1
$\begingroup$

Legend wants that Carl Friedrich Gauss discover the formula

$$ \sum_{i=1}^n i = \dfrac{n(n+1)}{2} $$ when he was six. Not surprising, since gaussing, ehm, guessing "Gauss" when trying to remember who found a certain result has a non trivial probability of success...

$\endgroup$
1
  • 2
    $\begingroup$ Ah! Interesting that I got a down vote. He asked if we heard of those formula's. I pointed out this formula was (probably) discovered by Gauss. Yes, it is not the 'exact' formula he wrote in the question, but it is the only brick needed to prove it (as all the others pointed out). Also, this is the only 'formula' people remember and use. Cheers. $\endgroup$
    – bartgol
    Aug 7, 2015 at 15:23
1
$\begingroup$

Ah man. I can't believe I'm late to this party. I discovered this as well a lot of years ago and came up with my own set of proofs.

I noticed that: $$1 + 2 + .. + (n -1) + n + (n - 1) + ... + 2 + 1 = n^2$$ (which is the same thing that you have)

Proof by Induction:

Base case: For n = 1:

$LHS = 1 = 1^2 = RHS$

Assuming that it is true for an integer $k > 1$: (i.e. $1 + 2 + ... + (k - 1) + k + (k -1 ) + ... + k = k^2$)

The case for k + 1 becomes:

$LHS = 1 + 2 + ... + k + (k + 1) + k + ... + 2 + 1$

$ = k^2 + (k + 1) + k$ (using the induction hypothesis)

$ = (k + 1)^2 = RHS $

$\endgroup$
0
$\begingroup$

In Zeilberger fashion: Plug in $n=2, 3$ into the LHS to get $4, 9$. Fit a quadratic to that and get $n^2$. Then to complete the proof, simply note

$$ \left(\sum^{n+1}_{i = 1}i + \sum^{n}_{i=1}i\right) - \left(\sum^n_{i = 1}i + \sum^{n-1}_{i=1}i\right) = n+n+1 = (n+1)^2-n^2 $$

$\endgroup$
2
  • $\begingroup$ ...And why the downvote? It's perfectly rigourous. $\endgroup$
    – Argon
    Aug 7, 2015 at 20:00
  • $\begingroup$ arxiv.org/pdf/1105.0178.pdf $\endgroup$
    – Argon
    Mar 4 at 16:54
0
$\begingroup$

$$ \frac{(n-1) n}{2}+\frac{ n(n+1)}{2} = n^2.$$

$\endgroup$
1
  • $\begingroup$ This way of looking at it was fully explained in AJ Stas's answer. Read the other answers before answering. $\endgroup$
    – user26486
    Aug 7, 2015 at 4:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .