# Difference between revisions of "Triangular number"

The triangular numbers are the numbers $T_n$ which are the sum of the first $n$ natural numbers from $1$ to $n$.

## Definition

The $n^{th}$ triangular number is the sum of all natural numbers from one to n. That is, the $n^{th}$ triangle number is $1 +2+3 + 4............. +(n-1)+(n)$.

For example, the first few triangular numbers can be calculated by adding 1, 1+2, 1+2+3, ... etc. giving the first few triangular numbers to be $1, 3, 6, 10, 15, 21$.

A rather simple recursive definition can be found by noting that $T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n$.

They are called triangular because you can make a triangle out of dots, and the number of dots will be a triangular number: $[asy] int draw_triangle(pair start, int n) { real rowStart = start.x; for (int row=1; row<=n; ++row) { for (real j=rowStart; j<(rowStart+row); ++j) { draw((j, start.y - row), linewidth(3)); } rowStart -= 0.5; } return 0; } for (int n=1; n<5; ++n) { real value= n*(n+1)/2; draw_triangle((value+5,n),n); label( (string) value, (value+5, -2)); } [/asy]$

## Formula

Using the sum of an arithmetic series formula, a formula can be calculated for $T_n$:

$T_n =\sum_{k=1}^{n}k = 1 + 2 + \ldots + n = \frac{n(n+1)}2$

The formula for finding the $n^{th}$ triangular number can be written as $\dfrac{n(n+1)}{2}$.

It can also be expressed as the sum of the $n^{th}$ row in Pascal's Triangle and all the rows above it. Keep in mind that the triangle starts at Row 0.