Triangular number

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


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]


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.

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