# Proof for formula for amount of series $1+2+3+\ldots+n$?

How several means exist to pick a $2$ - component part out of an $n$ - component set?

On the one hand, you can pick the first component of the embeded in $n$ means, after that the 2nd component of the embeded in $n-1$ means, after that divide by $2$ due to the fact that no matter which you pick first and also which you pick 2nd. This offers $\frac{n(n-1)}{2}$ means.

On the various other hand, intend the $n$ components are $1, 2, 3, ... n$, and also intend the bigger of both components you pick is $j$. After that for every single $j$ in between $2$ and also $n$ there are $j-1$ feasible selections of the smaller sized of both components, which can be any one of $1, 2, ... j-1$. This offers $1 + 2 + ... + (n-1)$ means.

Given that both expressions over matter the very same point, they have to be equivalent. This is called the concept of double checking, and also it is just one of a combinatorialist's favored tools. A generalization of this argument permits one to reason the amount of the first $n$ squares, dices, 4th powers ...

Once you have a formula similar to this, you can confirm it by induction. Yet that asks the inquiry regarding just how you get such a formula. In this instance you might ask : (a) what's the "ordinary" term? and also (b) the amount of terms exist?

The most basic instance of this is called an *math development * or * (limited) math collection *. There are several, several, several evidence. A very easy one : write all the summands straight ; write them once more simply listed below, yet from appropriate to left currently (so $1$ is under $n$, $2$ is under $n-1$, etc). Add them up, and also identify just how it connects to the amount you are seeking.

**HINT ** Pair each summand $k$ with it's "representation" $n+1-k$. This is merely a distinct analog of the method of calculating the location of the triangular under the diagonal of a square by mirroring a subtriangle via the middle of the angled to create an $n$ by $n/2$ rectangular shape.

Like the similar evidence of Wilson's theorem, the method manipulates the presence of a nontrivial proportion. In Wilson's theory we manipulate the proportion $n \mapsto n^{-1}$ which exists as a result of the reality that ${\mathbb F}_p^*$ creates a team. Below we manipulate a representation via a line - a proportion that exists as a result of the *straight * nature of the trouble (which does not benefit nonlinear amounts, as an example $\sum n^2$). Proportions usually bring about classy evidence. One need to constantly seek natural proportions when first contemplating troubles.

Usually there are (Galois) concepts and also formulas for **summation in shut kind **, in example to the differential instance (Ritt, Kolchin, Risch et al.). A *really * wonderful determined intro can be located in the initial phase of Carsten Schneider's thesis Symbolic Summation in Difference Fields.

Another "image evidence" I simply considered ... yet without an image, given that I can not attract :

Suppose you intend to build up all the integers from 1 to n. After that attract n rows on a board, placed 1 device in the first row, 2 in the 2nd, and more. If you attract an appropriate triangular of elevation and also size n to attempt and also have this form, you will certainly remove fifty percent of each device on the angled. So allow's add every one of these up ; you have n ^ 2/ 2 devices inside the triangular, and also you have actually n/2 joins removed on the angled (there are n squares on the angled, and also fifty percent of every one is removed). Including these offers n ^ 2/ 2+n/2 = n (n+1)/ 2.

If you attract it out it makes even more feeling, and also I assume it's geometrically a little bit extra uncomplicated than the various other image evidence.

(Side note : I have no suggestion just how to write mathematics on this website ... I'll go get in touch with the meta, I'm certain there's something there concerning it.)

My favorite evidence is the one offered here on MathOverflow. I'm replicating the image below for very easy reference, yet complete debt mosts likely to Mariano Suárez-Alvarez for this solution.

Takes a little of considering it to see what's taking place, yet it's wonderful once you get it. Observe that if there are n rows of yellow discs, after that :

- there are a total amount of 1+2+...+n yellow discs ;
- every yellow disc represents an one-of-a-kind set of blue discs, and also the other way around ;
- there are ${n+1 \choose 2} = \frac 12 n(n+1)$ such sets.

Let $$S = 1 + 2 + ... + (n-1) + n.$$ Write it in reverse : $S = n + (n-1) + ... + 2 + 1.$ Add both formulas, term by term ; each term is $n+1,$ so $$2S = (n+1) + (n+1) + ... + (n+1) = n(n+1).$$ Separate by 2 : $$S = \frac{n(n+1)}{2}$$.

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