Question:
Expand the function $\frac{2}{(2x – 3)(2x+1)}$ in a series of powers of $x$ up to $x^2$. State the set of values of $x$ for which this expansion is valid.
I've come across this question and would like to ask how most of you would tackle it. I've never seen one of this form before as it's my first time tackling fractional or negative indices. Thus I kindly ask to correct me wherever I'm wrong in my attempt:
My Attempt:
I rewrote $\frac{2}{(2x – 3)(2x+1)}$ as $2(2x – 3)^{-1}(2x+1)^{-1}$ since it seemed to somehow make some more sense. I then followed the simple formula $$1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!}$$
And substituted it with one of my terms whilst leaving 2 outside: $$2[(2x-3)^{-1}] \equiv 2[1+(-1)(2x)+\frac{(-1)(-2)(2x)^2}{2(1)}]$$
As the question requests up to $x^2$. Once working all of that out I got:
$$2[1-2x-4x^2]$$
$\therefore$ The expansion is valid when $x$ is between $-\frac{1}{4}$ and $\frac{1}{4}$
And wrote it down as an inequality: $$-\frac{1}{4} < x < \frac{1}{4}$$
Second Attempt
I went ahead and worked it out using Partial Fractions. Seems to make much more sense.
$$\frac{2}{(2x – 3)(2x+1)} \equiv \frac{A}{(2x-3)}+\frac{B}{(2x+1)}$$
Rewrite as..
$$2 = A(2x+1)+B(2x-3)$$
$$2 = \frac{2}{(2x+1)}-\frac{2}{(2x-3)}$$
And once again rewrite as..
$$2(2x+1)^{-1} – 2(2x-3)^{-1}$$
$$2^{-1}(2x+1)^{-1}+6^{-1}(\frac{2}{3}x-1)^{-1}$$
From here I worked each expansion separately however got stuck at a certain point.
$$\frac{1}{2}[1+(-1)(2x)+\frac{(-1)(-2)(2x)^2}{2!} ] = \frac{1}{2}[1 – 2x – 4x^2 ]$$
And..
$$\frac{1}{6}[1 + (-1)(\frac{2}{3}x) + \frac{(-1)(-2)(\frac{2}{3}x)^2}{2!} ] = \frac{1}{2}[1 – \frac{2}{3}x – (\frac{2}{3}x)^2]$$
I'm not sure if I'm doing it right from here onwards
$$\frac{1}{2}[1-2x-4x^2] – \frac{1}{6}[1-\frac{2}{3}x-(\frac{2}{3}x^2)]$$
And my final answer..
$$\frac{1}{2} – x – 2x^2 – \frac{1}{6} – \frac{1}{9}x – \frac{2}{27}x^2 = \frac{2}{3} – \frac{8}{9}x – \frac{56x^2}{27}$$
I'm sorry if this is completely wrong, it's literally my first time trying it out however it makes sense to consult an experienced person before entering the topic thinking I understand it when I don't.
Best Answer
The Binomial Theorem for negative powers says that for $|x| < 1$ $$(1+x)^{-1} = 1 - x + x^2 + \mathcal{o}(x^2)$$
Therefore we have: $$\frac 2{(2x-3)(2x+1)} = \frac 1{2(2x-3)} - \frac 1{2(2x+1)} = -\frac 16\left(1-\frac 23x\right)^{-1} - \frac 12\left(1+2x \right )^{-1} = -\frac16\left(1 + \frac 23x + \frac 49x^2\right)-\frac 12\left(1 - 2x + 4x^2 \right ) = \boxed{-\frac 23 + \frac 89 x - \frac{56}{27}x^2}$$
This holds for $|x| < \frac12$.