Elements of Algebra

CHAP. XIII. OF ALGEBRA. 125 enable us to conclude with certainty, that we shall have, generally, for any negative power of $a+b$; $\frac{1}{(a+b)^m} = \frac{1}{a^m} - \frac{m.b}{a^{m+1}} + \frac{m.(m-1).b^2}{2.a^{m+2}} - \frac{m.(m-1).(m-2).b^3}{2.3.a^{m+3}}, \ldots$ And, by means of this formula, we may transform all such fractions into infinite series, substituting fractions also, or fractional exponents, for $m$, in order to express irrational quantities. 375. The following considerations will illustrate this subject still farther: for we have seen that, $\frac{1}{a+b} = \frac{1}{a} - \frac{b}{a^2} + \frac{b^2}{a^3} - \frac{b^3}{a^4} + \frac{b^4}{a^5} - \frac{b^5}{a^6} + \ldots$ If, therefore, we multiply this series by $a+b$, the product ought to be $=1$; and this is found to be true, as will be seen by performing the multiplication: $\frac{1}{a} - \frac{b}{a^2} + \frac{b^2}{a^3} - \frac{b^3}{a^4} + \frac{b^4}{a^5} - \frac{b^5}{a^6} + \ldots$ $a+b$ $1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{b^3}{a^3} + \frac{b^4}{a^4} - \frac{b^5}{a^5} + \ldots$ $+ \frac{b}{a} - \frac{b^2}{a^2} + \frac{b^3}{a^3} - \frac{b^4}{a^4} + \frac{b^5}{a^5} - \ldots$ where all the terms but the first cancel each other. 376. We have also found that $\frac{1}{(a+b)^2} = \frac{1}{a^2} - \frac{2b}{a^3} + \frac{3b^2}{a^4} - \frac{4b^3}{a^5} + \frac{5b^4}{a^6} - \frac{6b^5}{a^7}, \ldots$ And if we multiply this series by $(a+b)^2$, the product ought also to be equal to 1. Now, $(a+b)^2 = a^2 + 2ab + b^2$, and $\frac{1}{a^2} - \frac{2b}{a^3} + \frac{3b^2}{a^4} - \frac{4b^3}{a^5} + \frac{5b^4}{a^6} - \frac{6b^5}{a^7} + \ldots$ $a^2 + 2ab + b^2$ $1 - \frac{2b}{a} + \frac{3b^2}{a^2} - \frac{4b^3}{a^3} + \frac{5b^4}{a^4} - \frac{6b^5}{a^5} + \ldots$ $+ \frac{2b}{a} - \frac{4b^2}{a^2} + \frac{6b^3}{a^3} - \frac{8b^4}{a^4} + \frac{10b^5}{a^5} - \ldots$ $+ \frac{b^2}{a^2} - \frac{2b^3}{a^3} + \frac{3b^4}{a^4} - \frac{4b^5}{a^5} + \ldots$