THE ELEMENTS OF THE TRUE ARITHMETIC OF INFINITES, &c.

Thus, also, in the series $1+4+7+10+13$, &c. $=\frac{1+2}{1-2+1}$, the number of terms is $\frac{1}{1-1}$, and the last term is $\frac{1+2}{1-1}$. And $\frac{1}{1-1} \times \frac{1+2}{1-1}$ is equal to $\frac{1+2}{1-2+1}$.

Again, in the series $1+3+6+10+15$, &c. $=\frac{1}{1-3+3-1}$, the last term is the series $1+2+3+4+5$, &c. $=\frac{1}{1-2+1}$, and the number of terms is $\frac{1}{1-1}$, and $\frac{1}{1-2+1} \times \frac{1}{1-1} = \frac{1}{1-3+3-1}$.

And in the series $1+6+24+80$, &c. $=\frac{1}{1-6+12-8}$, the last term is $\frac{1-1}{1-6+12-8}$, and the number of terms is $\frac{1}{1-1}$, and $\frac{1}{1-1} \times \frac{1-1}{1-6+12-8} = \frac{1}{1-6+12-8}$.

PROP. VIII.

In every series of terms in arithmetical or geometrical progression a sequence where each term is found by multiplying the previous one by a constant, or in any progression in which the terms mutually exceed each other, the last term is equal to the first term, added to the second term, diminished by the first; added to the third term, diminished by the second; added to the fourth term, diminished by the third; and so on. And if the number of terms be infinite, the last term is equal to the series multiplied by $1-1$.

Let the terms, whatever the series may be, be represented by a, b, c, d, e, then
$a + b - a + c - b + d - c + e - d = a$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + b - a$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c - b$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + d - c$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + e - d$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = e$

But if the number of terms be infinite, viz. if the series be $a+b+c+d+e+f+g$, &c. ad infin. to infinity then this series multiplied by $1-1$, will be