If |x| > 1, then substituting this into the expansion, the powers of $x$ increasing would lead to infinitely large numbers.

However, if |x| ≤ 1, then higher powers of $x$ would lead to a number of a smaller size meaning the series would converge!

Fact: For any series of the form $(1+ax)^n, a, n \in \mathbb{R}$, the series is only valid when |ax| < 1.

Example: State the values for which the expansion of $\sqrt{1 + 8x}$ is valid.

Example: State the values of $x$ for which $\dfrac{4}{\sqrt[3]{1-7x}}$ is valid.

e.g**. Expand** $(3+2x)^{-2}$ up to and including the $x^3$ term. State the values for which this expression is valid. Note that the formula we were given only works for brackets of the form (1+ax)^n. If given a number other than 1, we must take out a factor so that the bracket reads $k(1+ax)^n$.

Remember, everything in the bracket has power -2

(Can expand using formula)

Q3 (Jun 2014, Q3) (i) Find the first three terms in the expansion of $(1-2x)^{-\frac{1}{2}}$ in ascending powers of $x$, where $|x| < \frac{1}{2}$.

(ii) Hence find the coefficient of $x^2$ in the expansion of $\dfrac{x+3}{\sqrt{1-2x}}$.

(Already expanded $\dfrac{1}{\sqrt{1-2x}}$ in part (i))

(Only asked for the coefficient)