Question

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

Answer 1

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 `1/(1-(x/2))^{3}=(1+(-x/2))^{-3}`, we can apply this result with `p=-3` and `y=-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

`1/(1-(x/2))^{3}=1+3/2 x+3/2 x^{2}+ 5/4 x^[3}+ 15/16 x^{4}+cdots`.

This converges for `|-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.