# How do you use the binomial series to expand 1/sqrt(4+x)?

Nov 24, 2016

The answer is $= \frac{1}{2} - \frac{x}{16} + \frac{3 {x}^{2}}{256} - \frac{5 {x}^{3}}{2048} + \ldots \ldots$

#### Explanation:

The binomial theorem is
${\left(a + b\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} b + \left(\begin{matrix}n \\ 3\end{matrix}\right) {a}^{n - 2} {b}^{2} + \left(\begin{matrix}n \\ 4\end{matrix}\right) {a}^{n - 3} {b}^{3} + \ldots \ldots$

Where ((n),(p))=(n!)/((n-p)!p!)

Let's rewrite $\frac{1}{\sqrt{4 + x}}$

as $\frac{1}{2 \sqrt{1 + \frac{x}{4}}} = \frac{1}{2} {\left(1 + \frac{x}{4}\right)}^{- \frac{1}{2}}$

Now we can use the binomial theorem,

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

$= \frac{1}{2} \left(1 - \frac{1}{2} \cdot \frac{x}{4} + \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot {x}^{2} / 16 - \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{5}{2} \cdot \frac{1}{6} \cdot {x}^{3} / 64\right)$

$= \frac{1}{2} - \frac{x}{16} + \frac{3 {x}^{2}}{256} - \frac{5 {x}^{3}}{2048} + \ldots \ldots$