# How do you use the binomial series to expand f(x) = (6-x)^-3?

Apr 17, 2018

$\frac{1}{216} - \frac{1}{432} x + \frac{1}{1296} {x}^{2} - \frac{5}{23328} {x}^{3} + \frac{5}{93312} {x}^{4} - \frac{7}{559872} {x}^{5} + \ldots$

#### Explanation:

The binomial expansion when $n$ is fractional or negative is a little different in form than when $n$ is positive. We are not able to use the standard method of ${\textcolor{w h i t e}{0}}^{n} {C}_{r}$, since this is only valid for positive integers. Notice also that we do not have a finite number of terms. The number of terms required would have to be specified.

The form we will use is:

(1+x)^n-= 1+nx+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+..........

We have ${\left(6 + x\right)}^{-} 3$, we need to manipulate this to get the required form:

$\frac{1}{6} ^ 3 {\left(1 + \frac{x}{6}\right)}^{-} 3$

$1 + \left(- 3\right) \left(\frac{x}{6}\right) + \frac{\left(- 3\right) \left(- 4\right)}{2} {\left(\frac{x}{6}\right)}^{2} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right)}{6} {\left(\frac{x}{6}\right)}^{3} \to$

$+ \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right) \left(- 6\right)}{24} {\left(\frac{x}{6}\right)}^{4} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right) \left(- 6\right) \left(- 7\right)}{120} {\left(\frac{x}{6}\right)}^{5}$

$\frac{1}{216} \left(1 - \frac{1}{2} x + \frac{1}{6} {x}^{2} - \frac{5}{108} {x}^{3} + \frac{5}{432} {x}^{4} - \frac{7}{2592} {x}^{5} + \ldots\right)$

$\frac{1}{216} - \frac{1}{432} x + \frac{1}{1296} {x}^{2} - \frac{5}{23328} {x}^{3} + \frac{5}{93312} {x}^{4} - \frac{7}{559872} {x}^{5} + \ldots$

These series are always infinite, unless $| x | < 1$, or in this case $| \frac{x}{6} | < 1$