# How do you use the binomial series to expand  1/((2+x)^3)?

Feb 13, 2018

The binomial expansion is
$= \frac{1}{8} \left(1 - \frac{3}{2} x + \frac{3}{2} {x}^{2} - \frac{5}{4} {x}^{3} + \ldots\right)$ for $| x | < 2$ and

= 1/x^3 (1-6/x+24/x^2- 80/x^3+...) for $| x | > 2$

#### Explanation:

One important point to remember is that if the index $n$ is not a positive integer, the binomial expansion leads to an infinite series. In such a situation, it is important to keep in mind that the infinite binomial series for ${\left(1 + x\right)}^{n}$ converges only if $| x | < 1$.

For an arbitrary $n$, the binomial expansion for ${\left(1 + x\right)}^{n}$ is given by

1+n x+ {n(n-1)}/{2!} x^2+ {n(n-1)(n-2)}/{3!} x^3 +...

So we first recast the expression that we want to expand binomially in the form ${\left(1 + x\right)}^{n}$

Here we first use
$\frac{1}{{\left(2 + x\right)}^{3}} = \frac{1}{2} ^ 3 {\left(1 + \frac{x}{2}\right)}^{- 3}$

and then use the binomial expansion for $n = - 3$ to get

1/{(2+x)^3} = 1/2^3 (1+(-3) x/2 + {(-3)(-4)}/{2!} (x/2)^2 + {(-3)(-4)(-5)}/{3!} (x/2)^3 +.... )
$= \frac{1}{8} \left(1 - \frac{3}{2} x + \frac{3}{2} {x}^{2} - \frac{5}{4} {x}^{3} + \ldots\right)$

Of course this infinite series only converges for $| \frac{x}{2} | < 1$, i. e. $- 2 < x < 2$.

If $| \frac{x}{2} | > 1$, this series will diverge - rendering it useless! What saves the day is that in this case, $| \frac{2}{x} | < 1$ and so, we can still get a binomial expansion, but thus time we must start from
$\frac{1}{{\left(2 + x\right)}^{3}} = \frac{1}{x} ^ 3 {\left(1 + \frac{2}{x}\right)}^{- 3}$
and continue with

1/{(2+x)^3} = 1/x^3 (1+(-3) 2/x + {(-3)(-4)}/{2!} (2/x)^2 + {(-3)(-4)(-5)}/{3!} (2/x)^3 +.... )
$= \frac{1}{x} ^ 3 \left(1 - \frac{6}{x} + \frac{24}{x} ^ 2 - \frac{80}{x} ^ 3 + \ldots\right)$