I am trying to understand proofs of the $1+2+3+4+ \cdots$ series.
I'm puzzled by the 3rd point of this post where it is solved by binomial coefficient
https://math.stackexchange.com/a/2288/777575
He equates ${n+1 \choose 2}= \frac 12 n(n+1)$ such pairs. Which is true, but what I am confused about is how did he arrive at $\frac 12 n(n+1)$ based on the info of the post: is there a way that I am not seeing?
I'm not talking about reaching the formula through other methods not relevant to how it is solved in the post eg square numbers divided by $2$ etc.
We can count them: Say our set is ${1,\dots,n}$.
The $2$-subsets are $\underbrace{{1,2},{1,3},\dots,{1,n}}{(n-1)}$ and $\underbrace{{2,3},{2,4},\dots,{2,n}}{(n-2)}$ and $\underbrace{{3,4},{3,5},\dots,{3,n}}_{(n-3)}$ and $\dots$ and finally $\underbrace{{n-1,n}}_1$.
That is, there are:
$$\dbinom{n}{2}=(n-1)+(n-2)+(n-3)+\dots+(n-n)=\frac{n(n-1)}{2}$$
Many such subsets.
– Vepir May 29 '20 at 09:14