Elements of Algebra

CHAP. V. OF ALGEBRA. 143 Indices, 1 2 3 4 5 6 7 8 9, \ldots Arith. Prog. 1, 4, 7, 10, 13, 16, 19, 22, 25, \ldots Pentagon, 1, 5, 12, 22, 35, 51, 70, 92, 117, \ldots the indices shewing the side of each pentagon. 433. It follows from this, that if we make the side $n$, the pentagonal number will be $\frac{3n^2-n}{2} = \frac{n(3n-1)}{2}$. Let, for example, $n = 7$, the pentagon will be 70; and if the pentagon, whose side is 100, be required, we make $n = 100$, and obtain 14950 for the number sought. 434. If we suppose the difference to be 4, we arrive at hexagonal numbers, as we see by the following progressions: Indices, 1 2 3 4 5 6 7 8 9, \ldots Arith. Prog. 1, 5, 9, 13, 17, 21, 25, 29, 33, \ldots Hexagon, 1, 6, 15, 28, 45, 66, 91, 120, 153, \ldots where the indices still shew the side of each hexagon. 435. So that when the side is $n$, the hexagonal number is $2n^2 - n = n(2n-1)$; and we have farther to remark, that all the hexagonal numbers are also triangular; since, if we take of these last the first, the third, the fifth, \ldots we have precisely the series of hexagons. 436. In the same manner, we may find the numbers which are heptagonal, octagonal, \ldots It will be sufficient therefore to exhibit the following Table of formulæ for all numbers that are comprehended under the general name of polygonal numbers. Supposing the side to be represented by $n$, we have for the Triangle $\frac{n^2+n}{2} = \frac{n(n+1)}{2}$. Square $\frac{2n^2+0n}{2} = n^2$. v-gon $\frac{3n^2-n}{2} = \frac{n(3n-1)}{2}$. vi-gon $\frac{4n^2-2n}{2} = 2n^2-n = n(2n-1)$. vii-gon $\frac{5n^2-3n}{2} = \frac{n(5n-3)}{2}$. of these figures into triangles might furnish matter for many curious considerations, and for elegant transformations of the general formulæ, by which the polygonal numbers are expressed in this chapter; but it is unnecessary to dwell on them at present. F. T.