Page:Elementary algebra (1896).djvu/367

Rh 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

$$\left ( - 1 \right )^{r- 1} \frac{1\cdot3\cdot5\cdots(2r-3)}{{2^r} {r!}} x^r$$

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

The $$(r+1)$$th term

$$ \begin{align} &= \frac{(- 3)(- 4)(- 5) \dots (- 3 - r + 1)} {r!} (-x)^{r} \\ &= -a^{2r} \frac{(- 3)(- 4)(- 5) \dots (r + 2)} {r!}(-1)^r(-x)^{r} \\ &= (-1)^r \frac {3\cdot4\cdot5\cdots (r +2)}{1\cdot2\cdot3\cdots r}x^r \\ &= \frac {(r+1)(r+2)}{1\cdot2}x^r \\ \end{align} $$

by removing like factors from the numerator and denominator.

$$ \begin{align} ( 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{align} $$

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

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

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

= 5.01329, to five places of decimals.