Reduce formular to triangular number

Question

This formula is given:

$$ n + (n-1) + (n-2) + (n-3) + (n-4) + \ldots + 1 \geq 100. $$

In the textbook it is written, the above is reduced using the triangular formula to

$$ n (n+1) / 2 \geq 100. $$

But why can we reduce it? The general triangular formula is:

$$ T_n = \sum^n_{k=1} k = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}. $$

In my case the formula would look like this:

$$ T_n = \sum^1_{k=n} k = n + (n-1) + (n-2) + \ldots + 1 = \frac{n(n+1)}{2}. $$

Can I just say that it is equivalent?

Answer 1

Addition is commutative, so it doesn't matter in what order a finite set of numbers is added. What matters is the multiset of terms. In both cases in your question the multiset is

$$ \{1, 2, \dots, n-1, n\}. $$

Answer 2

Addition is commutative, so $1+2+\ldots+(n-1)+n = n+(n-1)+\ldots + 2+1$.