Art of Problem Solving

Triangular number: Difference between revisions

No edit summary
No edit summary
Line 8: Line 8:
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;

Revision as of 09:12, 14 October 2018

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 ton. 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.

   }
   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.



This article is a stub. Help us out by expanding it.