Difference between revisions of "Triangular number"

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.

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

@@ Line 1: / Line 1: @@
 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>.
+==Definition==
+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
+<math>1+2+3+4+\ldots+(n-1)+(n)</math>.
+For example, the first few triangular numbers can be calculated by adding
+, 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;
+  }
+  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 <math>T_n</math>:
@@ Line 5: / Line 43: @@
 :<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>.
-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.
+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}}

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Difference between revisions of "Triangular number"

Latest revision as of 09:48, 13 August 2024

Definition

Formula