# triangular numbers

The *triangular numbers ^{}* are defined by the series

$${t}_{n}=\sum _{i=1}^{n}i$$ |

That is, the $n$th triangular number is simply the sum of the first $n$ natural numbers^{}. The first few triangular numbers are

$$1,3,6,10,15,21,28,\mathrm{\dots}$$ |

The name triangular number comes from the fact that the summation defining ${t}_{n}$ can be visualized as the number of dots in

$$\begin{array}{cccccccc}\hfill \bullet \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \bullet \hfill & \hfill \hfill & \hfill \hfill \\ \hfill \mathrm{\vdots}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \mathrm{\vdots}\hfill & \hfill \mathrm{\ddots}\hfill & \hfill \hfill \end{array}$$ |

where the number of rows is equal to $n$.

The closed-form for the triangular numbers is

$$t(n)=\frac{n(n+1)}{2}$$ |

Legend has it that a grammar-school-aged Gauss was told by his teacher to sum up all the numbers from 1 to 100. He reasoned that each number $i$ could be paired up with $101-i$, to form a sum of $101$, and if this was done $100$ times, it would result in twice the actual sum (since each number would get used twice due to the pairing). Hence, the sum would be

$$1+2+3+\mathrm{\cdots}+100=\frac{100(101)}{2}$$ |

The same line of reasoning works to give us the closed form for any $n$.

Another way to derive the closed form is to assume that the $n$th triangular number is less than or equal to the $n$th square (that is, each row is less than or equal to $n$, so the sum of all rows must be less than or equal to $n\cdot n$ or ${n}^{2}$), and then use the first few triangular numbers to solve the general 2nd degree polynomial $A{n}^{2}+Bn+C$ for $A$, $B$, and $C$. This leads to $A=1/2$, $B=1/2$, and $C=0$, which is the same as the above formula^{} for $t(n)$.

Title | triangular numbers |
---|---|

Canonical name | TriangularNumbers |

Date of creation | 2013-03-22 12:16:08 |

Last modified on | 2013-03-22 12:16:08 |

Owner | akrowne (2) |

Last modified by | akrowne (2) |

Numerical id | 6 |

Author | akrowne (2) |

Entry type | Definition |

Classification | msc 11A99 |

Classification | msc 40-00 |