The triangular numbers are defined by the series

$$ t_n = \sum_{i=1}^n i $$

That is, the $n$ th triangular number is simply the sum of the first $n$ natural numbers. The first few triangular numbers are

$$ 1, 3, 6, 10, 15, 21, 28, \ldots $$

The name triangular number comes from the fact that the summation defining $t_n$ can be visualized as the number of dots in

$$ \begin{matrix} \bullet & & & & & & & \\ \bullet & \bullet & & & & & & \\ \bullet & \bullet & \bullet & & & & & \\ \bullet & \bullet & \bullet & \bullet & & & & \\ \bullet & \bullet & \bullet & \bullet & \bullet & & & \\ \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & & \\ \vdots & & & & & \vdots & \ddots & \end{matrix} $$

where the number of rows is equal to $n$ .

The closed-form for the triangular numbers is

$$ t(n) = \frac{n(n+1)}{2} $$

Legend has it that a grammar-school-aged Gauss was told by his teacher to sum up all the numbers from 1 to 100. He reasoned that each number $i$ could be paired up with $101-i$ , to form a sum of $101$ , and if this was done $100$ times, it would result in twice the actual sum (since each number would get used twice due to the pairing). Hence, the sum would be

$$ 1+2+3+\cdots+100 = \frac{100(101)}{2} $$

The same line of reasoning works to give us the closed form for any $n$ .

Another way to derive the closed form is to assume that the $n$ th triangular number is less than or equal to the $n$ th square (that is, each row is less than or equal to $n$ , so the sum of all rows must be less than or equal to $n\cdot n$ or $n^2$ ), and then use the first few triangular numbers to solve the general 2nd degree polynomial $An^2 + Bn + C$ for $A$ , $B$ , and $C$ . This leads to $A=1/2$ , $B=1/2$ , and $C=0$ , which is the same as the above formula for $t(n)$ .