Elements of Algebra

144 ELEMENTS SECT. III. VIII-gon - $\frac{6n^2-4n}{2} = 3n^2 - 2n = n(3n-2)$. IX-gon - $\frac{7n^2-5n}{2} = \frac{n(7n-5)}{2}$. X-gon - $\frac{8n^2-6n}{2} = 4n^2 - 3n = n(4n-3)$. XI-gon - $\frac{9n^2-7n}{2} = \frac{n(9n-7)}{2}$. XII-gon - $\frac{10n^2-8n}{2} = 5n^2 - 4n = n(5n-4)$. XX-gon - $\frac{18n^2-16n}{2} = 9n^2 - 8n = n(9n-8)$. XXV-gon - $\frac{23n^2-21n}{2} = \frac{n(23n-21)}{2}$. m-gon - $\frac{(m-2)n^2-(m-4)n}{2}$*. 437. So that the side being $n$, the $m$-gonal number will be represented by $\frac{(m-2)n^2-(m-4)n}{2}$; whence we may deduce all the possible polygonal numbers which have the side $n$. Thus, for example, if the bigonal numbers were required, we should have $m=2$, and consequently the number sought $= n$; that is to say, the bigonal numbers are the natural numbers, 1, 2, 3, &c.* If we make $m=3$, we have $\frac{n^2+n}{2}$ for the triangular number required. If we make $m=4$, we have the square number $n^2$, &c. 438. To illustrate this rule by examples, suppose that the xxv-gonal number, whose side is 36, were required; we * The general expression for the $m$-gonal number is easily derived from the summation of an arithmetical progression, whose first term is 1, common difference $d$, and number of terms $n$; as in the following series; viz. $1+(1+d)+(1+2d)+$, &c. $(1+(n-1)\cdot d)$, the sum of which is expressed by $\frac{(2+(n-1)\cdot d)n}{2}$; but in all cases $d=m-2$, therefore substituting this value for $d$, the expression becomes $\frac{2n+(n^2-n)\cdot(m-2)}{2} = \frac{(m-2)n^2-(m-4)n}{2}$ as in the formula.