Summation of the $n$ natural numbers from $1$ to $n$ can be calculated using this formula:
$$ \sum_{i=1}^{n}i = \frac{n(n+1)}{2} $$
and this formula is the same formula used for calculating the $n$th triangular number:
$$ T_n = \frac{n(n+1)}{2} $$
For example:
- $ T_1 = \frac{1(1+1)}{2} = 1 $
- $ T_2 = \frac{2(2+1)}{2} = 1 + 2 = 3 $
- $ T_3 = \frac{3(3+1)}{2} = 1 + 2 + 3 = 6 $