Elements of Algebra

CHAP. XI. OF ALGEBRA. 167 $1$ - - - $\frac{(1 \times 3)-1}{2} = 1$ $1+3$ - - - $\frac{(3 \times 3)-1}{2} = 4$ $1+3+9$ - - - $\frac{(3 \times 9)-1}{2} = 13$ $1+3+9+27$ - - $\frac{(3 \times 27)-1}{2} = 40$ $1+3+9+27+81$ - $\frac{(3 \times 81)-1}{2} = 121$. 514. Let us now suppose, generally, the first term to be $a$, the ratio $b$, the number of terms $n$, and their sum $s$, so that $s = a + ab + ab^2 + ab^3 + ab^4 + \dots ab^{n-1}$. If we multiply by $b$, we have $bs = ab + ab^2 + ab^3 + ab^4 + ab^5 + \dots ab^n$, and taking the difference between this and the above equation, there remains $(b-1)s = ab^n - a$; whence we easily deduce the sum required $s = \frac{a.(b^n-1)}{b-1}$. Consequently, the sum of any geometrical progression is found, by multiplying the last term by the ratio, or exponent of the progression, and dividing the difference between this product and the first term, by the difference between $1$ and the ratio. 515. Let there be a geometrical progression of seven terms, of which the first is $3$; and let the ratio be $2$: we shall then have $a=3$, $b=2$, and $n=7$; therefore the last term is $3 \times 2^6$, or $3 \times 64, = 192$; and the whole progression will be $3, 6, 12, 24, 48, 96, 192$. Farther, if we multiply the last term $192$ by the ratio $2$, we have $384$; subtracting the first term, there remains $381$; and dividing this by $b-1$, or by $1$, we have $381$ for the sum of the whole progression. 516. Again, let there be a geometrical progression of six terms, of which the first is $4$; and let the ratio be $\frac{3}{2}$: then the progression is $4, 6, 9, \frac{27}{2}, \frac{81}{4}, \frac{243}{8}$. If we multiply the last term by the ratio, we shall have $\frac{729}{16}$; and subtracting the first term $= \frac{64}{16}$, the remainder is $\frac{665}{16}$; which, divided by $b-1 = \frac{1}{2}$, gives $\frac{665}{8} = 83\frac{1}{8}$ for the sum of the series.