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_ + _x_² --------------- 1 + 2_x_ + _x_² 1 + 2_x_ + _x_² 1 + _x_ --------------- 1 + 2_x_ + _x_² _x_ + 2_x_² + _x_³ ----------------------- 1 + 3_x_ + 3_x_² + _x_³ Here are the second and third powers of 1 + _x_: the fourth, we can tell beforehand from the table, must be 1 + 4_x_ + 6_x_² + 4_x_³ + _x_⁴; and so on. Hence we have (1 + _x_)ⁿ = 0ₙ + 1ₙ_x_ + 2ₙ_x_² + 3ₙ_x_³ + ... + _n_ₙ_x_ⁿ which is usually written with the symbols 0ₙ, 1ₙ, &c. at length, thus, _n_ - 1 _n_ - 1 _n_ - 2 (1 + _x_)ⁿ = 1 + _nx_ + _n_-------_x_² + _n_------- -------_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 (_x_+_a_)ⁿ = _x_ⁿ+1ₙ_ax_ⁿ⁻¹+2ₙ_a_²_x_ⁿ⁻²+3ₙ_a_³_x_ⁿ⁻³+... +_n_ₙ_a_ⁿ 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_ _a_² _a_³ _a_⁴ 1 2_a_ 3_a_² 4_a_³ 1 3_a_ 6_a_² 1 4_a_ 1 where the oblique columns 1 + _a_, 1 + 2_a_ + _a_², 1 + 3_a_ + 3_a_² + _a_³, &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_ × 4_a_, _p_ × 6_a_², &c. at the bottom of the several columns; and if we had written at the top _x_⁴, _x_³, _x_², _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. Suppose we follow this mode of forming _p_(_x_ + _a_)³ + _q_(_x_ + _a_)² + _r_(_x_ + _a_) + _s_.