Background
Lately I've been self-studying Tom M. Apostol's Vol. 1 Calculus to make my understanding of the subject more rigorous after taking the actual class. I came across a proof for what the sum of the squares of the first $n$ natural numbers was – $$\sum_{i=1}^n i^2$$ and how it was equal to
$$\frac {n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\,.$$
In short, this involved the notion of a telescoping series, but I'd rather not go too deeply into the specifics. Rather, my interest turned to that of the sum of the first $n$ natural numbers, which was involved in the aforementioned demonstration.
Motivation
I'm sure plenty of you all have seen at least one proof of the following equality –
$$\sum_{i=1}^n i = 1 + 2 + \cdots + (n – 1) + n = \frac{n(n+1)}{2}\tag{1}\label{1}\\$$
Examples of these include a visual proof, proof by induction, etc. The purpose of this post is to explain my proof, whether it is valid, and how I could improve it if so. (Feel free to also critique my notation, I'd appreciate it.)
If done correctly, we should be able to find what the sum is equal to without making what I believe to be huge intuitive leaps.
Proof
Starting with \eqref{1}, we set the sum equal to a new variable $k$.
$$1 + 2 + \cdots + (n – 1) + n = k\tag{2}\label{2}\\$$
From here, one can realize that in some sense all the terms leading up to $n$ are fairly "close" in value to it; in other words, they can be put into the form $(n-a)$. As a result, \eqref{2} becomes –
$$(n – (n – 1)) + (n – (n – 2)) + \cdots + (n – 2) + (n – 1) + n$$
Rearranging the many $n$ together as well as the remaining terms, we get –
$$\underbrace{n + n + \cdots + n}_{n\text{ times}} + [-1 – 2 – \cdots – (n – 2) – (n -1)]$$
$n$ added unto itself $n$ times is the definition of $n^2$. This and factoring out the negative from the remaining sum gives –
$$n^2 – [1 + 2 + \cdots + (n – 2) + (n -1)]$$
Recall that this is still equal to the original sum, $k$. Notice as well that the remaining sum is equal to $k$ minus the $n$ term. Our equation transforms into –
$$n^2 – (k – n) = k \tag{3}\label{3}\\$$
Solving for $k$ (the original sum) results in –
$$k = \frac{n^2 + n}{2} = \frac{n(n + 1)}{2} \tag{4}\label{4}\\$$
Hence proving \eqref{1}.
Is this logically sound, or did I mess up somewhere?
Best Answer
This is correct but can be made simpler:
$$S=1+2+3+\cdots(n-1)+n$$ and
$$S=n+(n-1)+(n-2)+\cdots2+1,$$
so that by addition
$$2S=(n+1)+(n+1)+(n+1)+\cdots(n+1)+(n+1)=n(n+1).$$