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

Oct 29, 2016

The answer is $= 4 \left(1 + \frac{x}{2} - {x}^{2} / 8 + {x}^{3} / 16 + \ldots .\right)$

#### Explanation:

Use ${\left(a + b\right)}^{n} = {a}^{n} + \left({\text{_1^n)a^(n-1)b+(}}_{2}^{n}\right) {a}^{n - 2} {b}^{2.} \ldots .$
$= {a}^{n} + n {a}^{n - 1} b + \frac{\left(n\right) \left(n - 1\right)}{1 \cdot 2} {a}^{n - 2} {b}^{2} + \frac{\left(n\right) \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2 \cdot 3} . {a}^{n - 3} {b}^{3.} \ldots$
If we use the binomial development, we get
$4 {\left(1 + x\right)}^{\frac{1}{2}} = 4 \left(1 + \frac{x}{2} + \frac{\left(\frac{1}{2}\right) \left(_ \frac{1}{2}\right)}{1 \cdot 2} {x}^{2} + \frac{\left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right)}{1 \cdot 2 \cdot 3} {x}^{3} + \ldots .\right)$

$= 4 \left(1 + \frac{x}{2} - {x}^{2} / 8 + {x}^{3} / 16 + \ldots .\right)$