How do I get the expansion of $(2x-1)^{-1} $ to the $x^3$ term

Question

How do I get the expansion of $(2x-1)^{-1}$ to the $x^3$ term

I don't really know how I have to use the formula here.

Do I have to use $$(1-2x)^{-1} = 1 + nx + \cdots$$

Answer 1

For $g \in \mathbb{R}$, the geometric series $$\displaystyle\sum\limits_{n=0}^{\infty} g^n$$

converges when $-1 < g < 1$ to

$$1+g+g^2+g^3 + \cdots = \frac{1}{1-g}$$

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Let $g=2x$, so the radius of convergence becomes $1/2$.

Now, whenever $-1/2 < x < 1/2$ we have

$$ 1 + 2x + 4x^2 + 8x^3 + \cdots = \frac{1}{1-2x}$$

Since $2x-1$ = $-(1-2x)$, we can get the expansion you want by simply multiplying by $-1$:

$$\frac{1}{2x-1} = -1 -2x - 4x^2 - 8x^3 - \cdots$$

The interval of convergence for this is also $-1/2 < x < 1/2$.

Answer 2

We know that

$$(1-x)^{-1}=1+x+x^2+x^3+...$$

Then

$$(2x-1)^{-1}=-(1-2x)^{-1}=-1-(2x)-(2x)^2-(2x)^3-...$$