- 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