Thus, in the row 9, under the column headed 4, we see 126, which is 9 × 8 × 7 × 6 ÷ (1 × 2 × 3 × 4), the number of ways in which 4 can be chosen out of 9, which we represent by 4-{9}.
If we add the several rows, we have 1 + 1 or 2, 1 + 2 + 1 or 2², next 1 + 3 + 3 + 1 or 2³, &c. which verify a theorem already announced; and the law of formation shews us that the several columns are formed thus:
| 1 1 | 1 2 1 | 1 3 3 1 | |||
|---|---|---|---|---|---|
| 1 1 | 1 2 1 | 1 3 3 1 | |||
| 1 2 1 | 1 3 3 1 | 1 4 6 4 1 | , &c. |
so that the sum in each row must be double of the sum in the preceding. But we can carry the consequences of this mode of formation further. If we make the powers of 1 + x by actual algebraical
multiplication, we see that the process makes the same oblique addition in the formation of the numerical multipliers of the powers of x.
- 1 + x
- 1 + x
- 1 + x
- x + x²
- 1 + 2x + x²
- 1 + 2x + x²
- 1 + x
- 1 + 2x + x²
- x + 2x² + x³
- 1 + 3x + 3x² + x³
Here are the second and third powers of 1 + x: the fourth, we can tell beforehand from the table, must be 1 + 4x + 6x² + 4x³ + x⁴; and so on. Hence we have
(1 + x)ⁿ = 0ₙ + 1ₙx + 2ₙx² + 3ₙx³ + ... + nₙxⁿ
which is usually written with the symbols 0ₙ, 1ₙ, &c. at length, thus,
| (1 + x )ⁿ = 1 + nx + n | n - 1 | x ² + n | n - 1 | n - 2 | x ³ + &c. | |
|---|---|---|---|---|---|---|
| 2 | 2 | 3 |
This is the simplest case of what in algebra is called the binomial theorem. If instead of 1 + x we use x + a, we get