TriangularNumbers 2002-02-02 23:23:46 2002-02-02 23:33:40 Definition \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} %\usepackage{psfrag} %\usepackage{graphicx} %\usepackage{xypic} The \emph{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 $t(n)$.