Elements of Algebra

114 ELEMENTS SECT. II. In this arrangement, we perceive that the numerators begin by the exponent of the power required, and that they diminish successively by unity; while the denominators follow in the natural order of the numbers, $1, 2, 3, 4$, &c. Now, the first coefficient being always $1$, the first fraction gives the second coefficient; the product of the first two fractions, multiplied together, represents the third coefficient; the product of the three first fractions represents the fourth coefficient, and so on. Thus, the 1st coefficient is $1$ $= 1$ 2d - - - - $\frac{7}{1}$ $= 7$ 3d - - - - $\frac{7 \cdot 6}{1 \cdot 2}$ $= 21$ 4th - - - - $\frac{7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3}$ $= 35$ 5th - - - - $\frac{7 \cdot 6 \cdot 5 \cdot 4}{1 \cdot 2 \cdot 3 \cdot 4}$ $= 35$ 6th - - - - $\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}$ $= 21$ 7th - - - - $\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}$ $= 7$ 8th - - - - $\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}$ $= 1$ 349. So that we have, for the second power, the fractions $\frac{2}{1}, \frac{1}{2}$; whence the first coefficient is $1$, the second $\frac{2}{1}=2$, and the third $2 \times \frac{1}{2}=1$. The third power furnishes the fractions $\frac{3}{1}, \frac{2}{2}, \frac{1}{3}$; wherefore the 1st coefficient $= 1$; 2d $=\frac{3}{1}=3$; 3d $=3 \cdot \frac{2}{2}=3$; and 4th $=\frac{3}{1} \cdot \frac{2}{2} \cdot \frac{1}{3}=1$. We have, for the fourth power, the fractions $\frac{4}{1}, \frac{3}{2}, \frac{2}{3}, \frac{1}{4}$; consequently, the 1st coefficient $= 1$; 2d $\frac{4}{1}=4$; 3d $\frac{4}{1} \cdot \frac{3}{2}=6$; 4th $\frac{4}{1} \cdot \frac{3}{2} \cdot \frac{2}{3}=4$; and 5th $\frac{4}{1} \cdot \frac{3}{2} \cdot \frac{2}{3} \cdot \frac{1}{4}=1$. 350. This rule evidently renders it unnecessary to find the coefficients of the preceding powers, as it enables us to discover immediately the coefficients which belong to any one proposed. Thus, for the tenth power, we write the fractions $\frac{10}{1}, \frac{9}{2}, \frac{8}{3}, \frac{7}{4}, \frac{6}{5}, \frac{5}{6}, \frac{4}{7}, \frac{3}{8}, \frac{2}{9}, \frac{1}{10}$, by means of which we find the