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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 ton.
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The <math>n^{th}</math> triangular number is the sum of all natural numbers from one to <math>n</math>.
 
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>.
 
<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  
 
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
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    <math>}
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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>.
 
−
 
 
−
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>
 
−
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;
 
   }
 
   }
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   label( (string) value, (value+5, -2));
 
   label( (string) value, (value+5, -2));
 
}
 
}
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</asy>
+
</asy></math>
 +
 
 
==Formula==
 
==Formula==
  

Revision as of 19:46, 15 July 2020

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. 

   $}
   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>$ (Error compiling LaTeX. Unknown error_msg)

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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