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

Mar 23, 2018

The answer is $= \frac{1}{8} - \frac{3}{16} x + \frac{3}{16} {x}^{2} - \frac{5}{32} {x}^{3} + o \left({x}^{3}\right)$

#### Explanation:

The binomial series is

${\left(a + b\right)}^{n} = {a}^{n} + n {a}^{n - 1} b + \frac{\left(n\right) \left(n - 1\right)}{1 \cdot 2} {a}^{n - 2} {b}^{2} + \frac{\left(n\right) \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2 \cdot 3} {a}^{n - 3} {b}^{3} + \ldots . .$

Here,

$a = 2$

$b = x$

and

$n = - 3$

Therefore,

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

$= {2}^{- 3} + \left(- 3\right) \cdot \left({2}^{- 4}\right) \cdot x + \frac{- 3 \cdot - 4}{2} \cdot {2}^{- 5} {x}^{2} + \frac{- 3 \cdot - 4 \cdot - 5}{6} {2}^{- 6} {x}^{3} + \ldots .$

$= \frac{1}{8} - \frac{3}{16} x + \frac{3}{16} {x}^{2} - \frac{5}{32} {x}^{3} + o \left({x}^{3}\right)$