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Difference between revisions of "Triangular number"

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==Definition==
 
==Definition==
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The <math>n^{th}</math> triangular number is the sum of all natural numbers from one to <math>n</math>.
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The <math>n^{th}</math> triangular number is the sum of all natural numbers from one to n.
 
That is, the <math>n^{th}</math> triangle number is  
 
That is, the <math>n^{th}</math> triangle number is  
−
<math>1 +2+3 + 4............. +(n-1)+(n)</math>.
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<math>1+2+3+4+\ldots+(n-1)+(n)</math>.
  
 
For example, the first few triangular numbers can be calculated by adding  
 
For example, the first few triangular numbers can be calculated by adding  
 
1, 1+2, 1+2+3, ... etc.  
 
1, 1+2, 1+2+3, ... etc.  
 +
giving the first few triangular numbers to be
 +
<math>1, 3, 6, 10, 15, 21</math>.
 +
 +
A rather simple recursive definition can be found by noting that <math>T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n</math>.
 +
 +
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;
 
     rowStart -= 0.5;

Latest revision as of 09:48, 13 August 2024

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

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.



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