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Hometriangular numbers

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# 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

$1,3,6,10,15,21,28,\ldots$ |

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

$\begin{matrix}\bullet&&&&&&&\\ \bullet&\bullet&&&&&&\\ \bullet&\bullet&\bullet&&&&&\\ \bullet&\bullet&\bullet&\bullet&&&&\\ \bullet&\bullet&\bullet&\bullet&\bullet&&&\\ \bullet&\bullet&\bullet&\bullet&\bullet&\bullet&&\\ \vdots&&&&&\vdots&\ddots&\end{matrix}$ |

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

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

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 $An^{2}+Bn+C$ 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 $t(n)$.

## Mathematics Subject Classification

11A99*no label found*40-00

*no label found*

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