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

Dec 17, 2015

$\frac{1}{{\left(1 + 2 x\right)}^{3}} = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} \left(k + 1\right) \left(k + 2\right) {2}^{k - 1} {x}^{k}$

$- \frac{1}{2} < \setminus x < \frac{1}{2}$

#### Explanation:

(1 + x)^n = 1 + nx + frac{n(n-1)}{2!} x^2 + frac{n(n-1)(n-2)}{3!} x^3 + ...

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

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

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

$= {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} \frac{\left(k + 1\right) \left(k + 2\right)}{2} {2}^{k} {x}^{k}$

$= {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} \left(k + 1\right) \left(k + 2\right) {2}^{k - 1} {x}^{k}$