Elements of Algebra

CHAP. V. OF ALGEBRA. 141 Side `.`, `..`, `...`, `....` Triangle { `.`, `..`, `...` 428. A question therefore presents itself here, which is, how to determine the triangle when the side is given? and, after what has been said, this may be easily resolved. For if the side be $n$, the triangle will be $1+2+3+4+ \ldots -n$. Now, the sum of this progression is $\frac{n^2+n}{2}$; consequently the value of the triangle is $\frac{n^2+n}{2}$*. Thus, if $\begin{cases} n=1, \\ n=2, \\ n=3, \\ n=4, \end{cases}$ the triangle is $\begin{cases} 1, \\ 3, \\ 6, \\ 10, \end{cases}$ and so on: and when $n=100$, the triangle will be $5050$. 429. This formula $\frac{n^2+n}{2}$ is called the general formula of triangular numbers; because by it we find the triangular number, or the triangle, which answers to any side indicated by $n$. This may be transformed into $\frac{n(n+1)}{2}$; which serves also to facilitate the calculation; since one of the two numbers $n$, or $n+1$, must always be an even number, and consequently divisible by $2$. So, if $n=12$, the triangle is $\frac{12 \times 13}{2} = 6 \times 13 = 78$; and if $n=15$, the triangle is $\frac{15 \times 16}{2} = 15 \times 8 = 120$, \&c. 430. Let us now suppose the difference to be $2$, and we shall have the following arithmetical progression: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, \ldots$ the sums of which, taking successively one, two, three, four terms, \&c. form the following series: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, \ldots$ * M. de Joncourt published at the Hague, in 1762, a Table of trigonal numbers answering to all the natural numbers from $1$ to $20000$; which Tables are found useful in facilitating a great number of arithmetical operations, as the author shews in a very long introduction. F. T.