Elements of Algebra

124 ELEMENTS SECT. II. us reduce this quantity also to an infinite series. For this purpose, we must suppose $n = -2$, and we shall first have, for the coefficients, $\frac{n}{1} = -\frac{2}{1}$; $\frac{n-1}{2} = -\frac{3}{2}$; $\frac{n-2}{3} = -\frac{4}{3}$; $\frac{n-3}{4} = -\frac{5}{4}$, \&c.; and, for the powers of $a$, we obtain $a^n = \frac{1}{a^2}$; $a^{n-1} = \frac{1}{a^3}$; $a^{n-2} = \frac{1}{a^4}$; $a^{n-3} = \frac{1}{a^5}$, \&c. We have therefore $(a+b)^{-2} = \frac{1}{(a+b)^2} = \frac{1}{a^2} - \frac{2.b}{1.a^3} + \frac{2.3.b^2}{1.2.a^4} - \frac{2.3.4.b^3}{1.2.3.a^5} + \frac{2.3.4.5.b^4}{1.2.3.4.a^6}, \ldots$. Now, $\frac{2}{1} = 2$; $\frac{2.3}{1.2} = 3$; $\frac{2.3.4}{1.2.3} = 4$; $\frac{2.3.4.5}{1.2.3.4} = 5$, \&c. and consequently, $\frac{1}{(a+b)^2} = \frac{1}{a^2} - 2\frac{b}{a^3} + 3\frac{b^2}{a^4} - 4\frac{b^3}{a^5} + 5\frac{b^4}{a^6} - 6\frac{b^5}{a^7} + 7\frac{b^6}{a^8}, \ldots$. 373. Let us proceed, and suppose $n = -3$, and we shall have a series expressing the value of $\frac{1}{(a+b)^3}$ or of $(a+b)^{-3}$. Here the coefficients will be $\frac{n}{1} = -\frac{3}{1}$; $\frac{n-1}{2} = -\frac{4}{2}$; $\frac{n-2}{3} = -\frac{5}{3}$, \&c. and the powers of $a$ become, $a^n = \frac{1}{a^3}$; $a^{n-1} = \frac{1}{a^4}$; $a^{n-2} = \frac{1}{a^5}$, \&c. which gives $\frac{1}{(a+b)^3} = \frac{1}{a^3} - \frac{3.b}{1.a^4} + \frac{3.4.b^2}{1.2.a^5} - \frac{3.4.5.b^3}{1.2.3.a^6} + \frac{3.4.5.6.b^4}{1.2.3.4.a^7} = \frac{1}{a^3} - 3\frac{b}{a^4} + 6\frac{b^2}{a^5} - 10\frac{b^3}{a^6} + 15\frac{b^4}{a^7} - 21\frac{b^5}{a^8} + 28\frac{b^6}{a^9}, \ldots$. If now we make $n = -4$; we shall have for the coefficients $\frac{n}{1} = -\frac{4}{1}$; $\frac{n-1}{2} = -\frac{5}{2}$; $\frac{n-2}{3} = -\frac{6}{3}$; $\frac{n-3}{4} = -\frac{7}{4}$, \&c. And for the powers, $a^n = \frac{1}{a^4}$; $a^{n-1} = \frac{1}{a^5}$; $a^{n-2} = \frac{1}{a^6}$; $a^{n-3} = \frac{1}{a^7}$; $a^{n-4} = \frac{1}{a^8}$, whence we obtain, $\frac{1}{(a+b)^4} = \frac{1}{a^4} - \frac{4b}{1a^5} + \frac{4.5.b^2}{1.2.a^6} - \frac{4.5.6.b^3}{1.2.3.a^7}, \ldots = \frac{1}{a^4} - 4\frac{b}{a^5} + 10\frac{b^2}{a^6} - 20\frac{b^3}{a^7} + 35\frac{b^4}{a^8} - 56\frac{b^5}{a^9} +, \ldots$. 374. The different cases that have been considered