It is plain that
rrrr - 1 = rrrr - rrr + rrr - rr + rr - r + r - 1
Now (54), rr-r = r(r-1), rrr -rr = rr(r-1), rrrr-rrr = rrr(r-1), and the above equation becomes rrrr -1 = rrr(r-1) + rr (r-1) + r (r-1) + r-1; which is (54) rrr + rr + r + 1 taken r-1 times. Hence, rrrr-1 divided by r-1 will give 1 + r + rr + rrr, the sum of the terms required. In this way may be proved the following series of equations:
| 1 + r | = | rr - 1 |
| r - 1 | ||
| 1 + r + rr | = | rrr - 1 |
| r - 1 | ||
| 1 + r + rr + rrr | = | rrrr - 1 |
| r - 1 | ||
| 1 + r + rr + rrr + rrrr | = | rrrrr - 1 |
| r - 1 |
If r be less than unity, in order to find 1 + r + rr + rrr, observe that
1 - rrrr = 1 - r + r - rr + rr - rrr + rrr - rrrr
= 1 - r + r(1 - r) + rr(1 - r) + rrr(1 - r);
whence, by similar reasoning, 1 + r + rr + rrr is found by dividing 1-rrrr by 1-r; and equations similar to these just given may be found, which are,
| 1 + r | = | 1 - rr |
| 1 - r | ||
| 1 + r + rr | = | 1 - rrr |
| 1 - r | ||
| 1 + r + rr + rrr | = | 1 - rrrr |
| 1 - r | ||
| 1 + r + rr + rrr + rrrr | = | 1 - rrrrr |
| 1 - r |
The rule is: To find the sum of n terms of the series, 1 + r + rr + &c., divide the difference between 1 and the (n + 1)ᵗʰ term by the difference between 1 and r.
194. This may be applied to finding the sum of any number of terms of a continued proportion. Let a, b, c, &c. be the terms of which it is required to sum four, that is, to find a + b + c + d, or (192) a + ar + arr + arrr, or (54) a(1 + r + rr + rrr), which (193) is