# Show the first three terms in the expansion of 1/(1-(x/2))^3 in ascending powers of x?

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Bill K. Share
Dec 31, 2017

It sounds like you are after the Taylor (or Maclaurin) series for this function centered at $x = 0$, which is really a calculus topic, but it can be answered here using Newton's generalized binomial theorem (to get a so-called "binomial series ").

#### Explanation:

Newton's generalized binomial theorem says that, if $p$ is any real number and $| y | < 1$, then:

(1+y)^{p}=1+py+(p(p-1))/(2!)y^{2}+(p(p-1)(p-2))/(3!)y^{3}+(p(p-1)(p-2)(p-3))/(4!)y^{4}+cdots

(this truncates to a finite sum that converges for all $y$ when $p$ is a non-negative integer.)

Since $\frac{1}{1 - \left(\frac{x}{2}\right)} ^ \left\{3\right\} = {\left(1 + \left(- \frac{x}{2}\right)\right)}^{- 3}$, we can apply this result with $p = - 3$ and $y = - \frac{x}{2}$ to get:

1/(1-(x/2))^{3}=1-3(-x/2)+((-3)(-4))/(2!)(-x/2)^{2}+((-3)(-4)(-5))/(3!)(-x/2)^{3}+((-3)(-4)(-5)(-6))/(4!)(-x/2)^{4}+cdots

Simplifying this gives

$\frac{1}{1 - \left(\frac{x}{2}\right)} ^ \left\{3\right\} = 1 + \frac{3}{2} x + \frac{3}{2} {x}^{2} + \frac{5}{4} {x}^{3} + \frac{15}{16} {x}^{4} + \cdots$.

This converges for $| - \frac{x}{2} | < 1$, which is equivalent to $| x | < 2$ and $- 2 < x < 2$.

Take the first three of these terms to get the answer that is requested.

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