Tomemos cuatro términos de la serie, 1, r, rr, etc. o,
1 + r + rr + rrr
Es evidente que 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 |
Si r es menor que la unidad, para hallar 1 + r + rr + rrr, obsérvese que
1 - rrrr = 1 - r + r - rr + rr