Difference between revisions of "Triangular number"

(Did some stuff, added some stuff, organized some stuff)
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The '''triangular numbers''' are the numbers <math>T_n</math> which are the sum of the first <math>n</math> [[natural number]]s from <math>1</math> to <math>n</math>.  
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The '''triangular numbers''' are the numbers <math>T_n</math> which are the sum of the first <math>n</math> [[natural number]]s from <math>1</math> to <math>n</math>.
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==Definition==
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The <math>n^{th}</math> triangular number is the sum of all natural numbers from one to n.
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That is, the <math>n^{th}</math> triangle number is
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<math>1 +2+3 + 4............. +(n-1)+(n)</math>.
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For example, the first few triangular numbers can be calculated by adding
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1, 1+2, 1+2+3, ... etc.
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giving the first few triangular numbers to be
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<math>1, 3, 6, 10, 15, 21</math>.
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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>.
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They are called triangular because you can make a triangle out of dots, and the number of dots will be a triangular number:
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<asy>
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int draw_triangle(pair start, int n)
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{
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  real rowStart = start.x;
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  for (int row=1; row<=n; ++row)
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  {
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    for (real j=rowStart; j<(rowStart+row); ++j)
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    {
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      draw((j, start.y - row), linewidth(3));
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    }
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    rowStart -= 0.5;
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  }
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  return 0;
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}
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for (int n=1; n<5; ++n)
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{
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  real value= n*(n+1)/2;
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  draw_triangle((value+5,n),n);
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  label( (string) value, (value+5, -2));
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}
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</asy>
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==Formula==
  
 
Using the sum of an [[arithmetic series]] formula, a formula can be calculated for <math>T_n</math>:
 
Using the sum of an [[arithmetic series]] formula, a formula can be calculated for <math>T_n</math>:
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:<math>T_n =\sum_{k=1}^{n}k = 1 + 2 + \ldots + n = \frac{n(n+1)}2</math>
 
:<math>T_n =\sum_{k=1}^{n}k = 1 + 2 + \ldots + n = \frac{n(n+1)}2</math>
  
For example, the <math>n^{th}</math> triangle number is <math>1 +2+3 + 4............. +(n-1)+(n)</math>
 
  
 
The formula for finding the <math>n^{th}</math> triangular number can be written as <math>\dfrac{n(n+1)}{2}</math>.
 
The formula for finding the <math>n^{th}</math> triangular number can be written as <math>\dfrac{n(n+1)}{2}</math>.
  
It can also be expressed as the sum of the <math>n^{th}</math> 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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It can also be expressed as the sum of the <math>n^{th}</math> row in [[Pascal's Triangle]] and all the rows above it. Keep in mind that the triangle starts at Row 0.
 +
 
  
Pascal's Triangle:
 
[[File: Pascal's Triangle.png]]
 
  
  
The rather simple recursive definition can be easily found by noting that <math>T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n</math>.
 
  
 
{{stub}}
 
{{stub}}

Revision as of 21:48, 13 February 2016

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.



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