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The sum of $1+2+3+ . . . +n = n(n+1)/2$, as I have checked carefully, but how can you prove this? I am determining the dimension of the space of $n x n$ matrices in the upper triangular form, and it is clear to me that the number of basis matrices has to be $1 + 2 + 3 + . . . +n$. I found in a book that the number of basis matrices for a U.T. matrix is $n(n+1)/2$, which matches my idea but it is certainly a much more elegant way to put it. But how do you go from one to the other?

Kevin Long
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j. clover
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    Have you actually tried searching this site and others? You will find many many many ... different approaches. – Daniel W. Farlow Sep 02 '16 at 18:33
  • Like a lot of facts in mathematics, this was discovered many times by many people. I'm a particular fan of the proof Gauss did as a youngster. ( http://mathandmultimedia.com/2010/09/15/sum-first-n-positive-integers/ ) – Sean Lake Sep 02 '16 at 18:35
  • But the area of the Trapezoid was known long before! – G Cab Sep 02 '16 at 18:41
  • One 'amusing' approach is $$ \sum_{k = 1}^{n}k = \sum_{k = 1}^{n}k^{\underline{1}} = \left.{k^{\underline{2}} \over 2}\right\vert_{\ 1}^{\ n + 1} = {\left(,n+1,\right)^{\underline{2}} - 1^{\underline{2}} \over 2} = {\left(,n+1,\right)n \over 2} $$ – Felix Marin Sep 02 '16 at 22:51

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First, you can use simple induction, or else the following trick with a slighly concealed induction too:

$$\begin{align*}&S=1&+&2&+&3&+\ldots&+&(n-1)&+&n\\&S=n&+&(n-1)&+&(n-2)&+\ldots&+&2&+&1\end{align*}$$

Sum up both expressions above summandwise and get:

$$2S=(n+1) + (n+1)+\ldots+(n+1)=n(n+1)$$

and we're done. Perhaps the last form is something similar to what legend has said about Gauss solving this as an elementary school pupil.

DonAntonio
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  • I always heard that story about Gauss, too, but I don't know if it's really true or not. +1. – MPW Sep 02 '16 at 19:11
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The sum $1+2+3+4+5+\dots+n$, when $n$ is even, can be written as

   1 +   2   +   3   + ... +   (n/2)
+  n + (n-1) + (n-2) + ... + (n/2 + 1)

The sum of each column is $n+1$. The there are $n/2$ columns. So the total is

$(n+1) \times \frac{n}{2} = \frac{n(n+1)}{2}$

tmlen
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In grade school they study math mountains.The bottom row of the seven math mountain is 1+6. Above this is 2+5 and above this is 3+4.. There are two ways to add the numbers in the seven math mountain: go up one side and down the other ((1+2+3 + 4+5+6) or realize that each layer of the mountain adds to 7, which is why it is called the 7 math mountain, and the count that there are three layers. Thus, the seven math mountain provides a wonderful opportunity for an elementary-school teacher to let the students make a discovery and to teach an elementary principle of mathematics: you can often learn something by doing something (in this case adding) in two different ways.

Airymouse
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