Page:Elementary algebra (1896).djvu/367

The number of factors in the numerator is $r$, and $r-1$ of these are negative; therefore, by taking -1 out of each of these negative factors, we may write the above expression

$\left(-1\right)^{r-1}{\frac {1\cdot 3\cdot 5\cdots (2r-3)}{{2^{r}}{r!}}}x^{r}$

Ex. 2. Find the general term in the expansion of $(1-x)^{-3}$.

The $(r+1)$th term

${\begin{aligned}&={\frac {(-3)(-4)(-5)\dots (-3-r+1)}{r!}}(-x)^{r}\\&=-a^{2r}{\frac {(-3)(-4)(-5)\dots (r+2)}{r!}}(-1)^{r}(-x)^{r}\\&=(-1)^{r}{\frac {3\cdot 4\cdot 5\cdots (r+2)}{1\cdot 2\cdot 3\cdots r}}x^{r}\\&={\frac {(r+1)(r+2)}{1\cdot 2}}x^{r}\\\end{aligned}}$

by removing like factors from the numerator and denominator.

${\begin{aligned}(1-x)^{-1}&=1+x+x^{2}+x^{3}+\cdots +x^{x}+\cdots .\\(1-x)^{-2}&=1+2x+3x^{2}+4x^{3}+\cdots +(r+1)x^{x}\\(1-x)^{-3}&=1+3x+6x^{2}+10x^{3}\cdots +{\frac {\left(r+1\right)\left(r+2\right)}{1.2}}x^{r}+\cdots .\\\end{aligned}}$

424. The following example illustrates a useful application of the Binomial Theorem.

Ex. Find the cube root of 126 to 5 places of decimals.

${\begin{aligned}\left(126\right)^{\frac {1}{3}}&=5(5^{3}+1)^{\frac {1}{3}}=5\left(1+{\frac {1}{5^{3}}}\right)^{\frac {1}{3}}\\&=5\left(1+{\frac {1}{3}}\cdot {\frac {1}{5^{3}}}-{\frac {1}{9}}\cdot {\frac {1}{5^{6}}}+{\frac {5}{81}}\cdot {\frac {1}{5^{9}}}-\cdots \right)\\&=5+{\frac {1}{3}}\cdot {\frac {1}{5^{2}}}-{\frac {1}{9}}\cdot {\frac {1}{5^{5}}}+{\frac {1}{81}}\cdot {\frac {1}{5^{7}}}-\cdots \\&=5+{\frac {1}{3}}\cdot {\frac {2^{2}}{10^{2}}}-{\frac {1}{9}}\cdot {\frac {2^{5}}{10^{5}}}+{\frac {1}{81}}\cdot {\frac {2^{7}}{10^{7}}}-\cdots \\&=5+{\frac {.04}{3}}-{\frac {.00032}{9}}+{\frac {.0000128}{81}}-\cdots \\&=5+.01333\cdots -.000035\cdots +\cdots \\\end{aligned}}$

= 5.01329, to five places of decimals.