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

Evidently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$.

Just how? What's the evidence? Or possibly it is self noticeable simply considering the above?

PS : This trouble is called "The amount of the first $n$ favorable integers".

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2019-05-09 11:19:32
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Answers: 7

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 ...

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2019-05-11 23:44:47
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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?

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2019-05-10 08:03:13
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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.

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2019-05-10 07:53:20
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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.

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2019-05-10 07:48:21
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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.)

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2019-05-10 07:29:15
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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 :

1. there are a total amount of 1+2+...+n yellow discs ;
2. every yellow disc represents an one-of-a-kind set of blue discs, and also the other way around ;
3. there are ${n+1 \choose 2} = \frac 12 n(n+1)$ such sets.
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2019-05-10 06:31:54
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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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2019-05-10 06:07:37
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