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

Nov 4, 2016

the answer is $f \left(x\right) = 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + 15 {x}^{4} + \ldots . .$

Explanation:

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

Applying this here with $a = 1$, $b = x$ and $n = - 3$
we get
$f \left(x\right) = {\left(1 + x\right)}^{-} 3 = 1 - 3 x + \frac{\left(- 3\right) \left(- 4\right)}{1 \cdot 2} {x}^{2} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right)}{1 \cdot 2 \cdot 3} {x}^{3} + \frac{\left(- 3\right) \left(- 4\right) \left(- 5\right) \left(- 6\right)}{1 \cdot 2 \cdot 3 \cdot 4} {x}^{4} + \ldots \ldots$
$= 1 - 3 x + 6 {x}^{2} - 10 {x}^{3} + 15 {x}^{4} + \ldots . .$