Elements of Algebra

CHAP. IV. OF ALGEBRA. 137 fore begin by determining the last term; which is found by the rule, in Articles 406 and 411, to be $5 + (31 \times 3) = 98$; after which the sum required is easily found to be $\frac{103 \times 32}{2}$ $= 103 \times 16$; whence we conclude that the horse costs $1648$ pence, or $6l. 17s. 4d$. 420. Generally, let the first term be $a$, the difference $d$, and the number of terms $n$; and let it be required to find, by means of these data, the sum of the whole progression. As the last term must be $a \pm (n-1)d$, the sum of the first and the last will be $2a \pm (n-1)d$; and multiplying this sum by the number of terms $n$, we have $2na \pm n(n-1)d$; the sum required therefore will be $na \pm \frac{n(n-1)d}{2}$. Now, this formula, if applied to the preceding example, or to $a = 5$, $d = 3$, and $n = 32$, gives $5 \times 32 + \frac{32 \cdot 31 \cdot 3}{2} = 160 + 1488 = 1648$; the same sum that we obtained before. 421. If it be required to add together all the natural numbers from $1$ to $n$, we have, for finding this sum, the first term $1$, the last term $n$, and the number of terms $n$; therefore the sum required is $\frac{n^2+n}{2} = \frac{n(n+1)}{2}$. If we make $n = 1766$, the sum of all the numbers, from $1$ to $1766$, will be $883$, or half the number of terms, multiplied by $1767 = 1560261$. 422. Let the progression of uneven numbers be proposed, such as $1, 3, 5, 7, \&c.$ continued to $n$ terms, and let the sum of it be required. Here the first term is $1$, the difference $2$, the number of terms $n$; the last term will therefore be $1 + (n-1)2 = 2n - 1$, and consequently the sum required $= n^2$. The whole therefore consists in multiplying the number of terms by itself; so that whatever number of terms of this progression we add together, the sum will be always a square, namely, the square of the number of terms; which we shall exemplify as follows: Indices, $1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9 \ 10, \&c.$ Progress. $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, \&c.$ Sum. $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \&c.$ 423. Let the first term be $1$, the difference $3$, and the number of terms $n$; we shall have the progression $1, 4, 7, 10, \&c.$ the last term of which will be $1+(n-1)3=3n-2$;