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

$\frac{1}{2 + x} ^ 2 = \frac{1}{4 + 4 x + {x}^{2}}$

#### Explanation:

The binomial expansion allows us to take something in the form

${\left(a + b\right)}^{n}$

and express it as a sum of terms:

${\left(a + b\right)}^{n} = \left({C}_{n , 0}\right) {a}^{n} {b}^{0} + \left({C}_{n , 1}\right) {a}^{n - 1} {b}^{1} + \ldots + \left({C}_{n , n}\right) {a}^{0} {b}^{n}$

We can do this for the denominator in the fraction:

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

and so:

$\frac{1}{2 + x} ^ 2 = \frac{1}{4 + 4 x + {x}^{2}}$

May 9, 2017

$\frac{1}{2 + x} ^ 2 = \frac{1}{4} - \frac{1}{4} x + \frac{3}{16} {x}^{2} - \frac{1}{8} {x}^{3} + \frac{5}{64} {x}^{4} - \frac{3}{64} {x}^{5} + \ldots \ldots \ldots \ldots \ldots$

#### Explanation:

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

Now Binomial expansion of ${\left(1 + a\right)}^{n}$

= $1 + n a + \frac{n \left(n - 1\right)}{1 \cdot 2} {a}^{2} + \frac{n \left(n - 1\right) \left(n - 2\right)}{1 \cdot 2 \cdot 3} {a}^{3} + \frac{n \left(n - 1\right) \left(n - 2\right) \left(n - 3\right)}{1 \cdot 2 \cdot 3 \cdot 4} {a}^{4} + \ldots \ldots \ldots \ldots .$

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

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

= $1 - x + \frac{3}{4} {x}^{2} - \frac{1}{2} {x}^{3} + \frac{5}{16} {x}^{4} - \frac{3}{16} {x}^{5} + \ldots \ldots \ldots \ldots \ldots$

and $\frac{1}{2 + x} ^ 2 = \frac{1}{4} \times \left(1 - x + \frac{3}{4} {x}^{2} - \frac{1}{2} {x}^{3} + \frac{5}{16} {x}^{4} - \frac{3}{16} {x}^{5} + \ldots \ldots \ldots \ldots \ldots\right)$

= $\frac{1}{4} - \frac{1}{4} x + \frac{3}{16} {x}^{2} - \frac{1}{8} {x}^{3} + \frac{5}{64} {x}^{4} - \frac{3}{64} {x}^{5} + \ldots \ldots \ldots \ldots \ldots$