(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 | x² + n | n - 1 | n - 2 | x³ + &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
(x + a)ⁿ = xⁿ + 1ₙaxⁿ⁻¹ + 2ₙa²xⁿ⁻² + 3ₙa³xⁿ⁻³ + ... + nₙaⁿ
We can make the same table in another form. If we take a row of ciphers beginning with unity, and setting down the first, add the next, and then the next, and so on, and then repeat the process with one step less, and then again with one step less, we have the following:
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 3 | 6 | 10 | 15 | ||
| 1 | 4 | 10 | 20 | |||
| 1 | 5 | 15 | ||||
| 1 | 6 | |||||
| 1 | ||||||
In the oblique columns we see 1 1, 1 2 1, 1 3 3 1, &c. the same as in the original table, and formed by the same additions. If, before making the additions, we had always multiplied by a, we should have got the several components of the powers of 1 + a, thus,
| 1 | 0 | 0 | 0 | 0 |
| 1 | a | a² | a³ | a⁴ |
| 1 | 2a | 3a² | 4a³ | |
| 1 | 3a | 6a² | ||
| 1 | 4a | |||
| 1 | ||||
where the oblique columns 1 + a, 1 + 2a + a², 1 + 3a + 3a² + a³, &c., give the several powers of 1 + a. If instead of beginning with 1, 0, 0, &c. we had begun with p, 0, 0, &c. we should have got p, p × 4a, p × 6a², &c. at the bottom of the several columns; and if we had written at the top x⁴, x³, x², x, 1, we should have had all the materials for forming p(x + a)⁴ by multiplying the terms at the top and bottom of each column together, and adding the results.