Binomial expansion for fractional and negative powers

For any rational $n$ and $|x| < 1$:

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

The pattern never changes. The exponent $n$ is an ordinary number, so substitute it and simplify.

Two things differ from AS Level:

• The series never stops, unless $n$ is a positive whole number. There is no last term, so the answer is quoted "up to and including the term in $x^3$".
• It converges only for $|x| < 1$. "Converges" means adding more and more terms gets closer and closer to one fixed value; "diverges" means it does not. Beyond that the terms grow instead of shrinking.

A quick sanity check with $n = -1$: the terms are $1, -x, x^2, -x^3, \dots$, which is the familiar geometric series. That series is valid only for $|x| < 1$, as you already know.

The series needs a 1 at the front of the bracket. Take $\left(2 - \tfrac{1}{2}x\right)^{-1}$ and factor out the 2:

$\left(2 - \tfrac{1}{2}x\right)^{-1} = \left[2\left(1 - \tfrac{1}{4}x\right)\right]^{-1} = 2^{-1}\left(1 - \tfrac{1}{4}x\right)^{-1}$

So the routine is:

1. Factor out the constant so the bracket starts with 1, and raise that constant to the power $n$.
2. Expand the bracket with the series, then multiply every term by the constant.

Validity uses the inner expression. Here it needs $\left|-\tfrac{1}{4}x\right| < 1$, which gives $|x| < 4$.

The forgotten step: the power $n$ applies to the constant too. The factor is $2^{-1}$, not $2$.

For $(2 + x)^{-\frac{1}{2}}$ the factor is $2^{-\frac{1}{2}} = \dfrac{1}{\sqrt{2}}$.