# Thread: Rational expression and binomial expansion

1. ## Rational expression and binomial expansion

Find the first four terms of the series expansion for the rational expression: 2+x / (1+4x)(1-3x)

In my attempt at this question I have simplified as partial fractions, but they don't look right, and also after using binomial expansion to find the first four terms, my terms are incorrect.

2. ## Re: Rational expression and binomial expansion

Sorry guys, already solved it myself

3. ## Re: Rational expression and binomial expansion

Originally Posted by Bernana
Find the first four terms of the series expansion for the rational expression: 2+x / (1+4x)(1-3x)

In my attempt at this question I have simplified as partial fractions, but they don't look right, and also after using binomial expansion to find the first four terms, my terms are incorrect.
\displaystyle \displaystyle \begin{align*} \frac{A}{1 + 4x} + \frac{B}{1 - 3x} &\equiv \frac{2 + x}{(1 + 4x)(1 - 3x)} \\ \frac{A(1 - 3x) + B(1 + 4x)}{(1 + 4x)(1 - 3x)} &\equiv \frac{2 + x}{(1 + 4x)(1 - 3x)} \\ A(1 - 3x) + B(1 + 4x) &\equiv 2 + x \end{align*}
Now let \displaystyle \displaystyle \begin{align*} x = \frac{1}{3} \end{align*} and we find \displaystyle \displaystyle \begin{align*} B\left[ 1 + 4 \left( \frac{1}{3} \right) \right] = 2 + \frac{1}{3} \implies \frac{7}{3}B = \frac{7}{3} \implies B = 1 \end{align*} and let \displaystyle \displaystyle \begin{align*} x = -\frac{1}{4} \end{align*} and we find \displaystyle \displaystyle \begin{align*} A \left[ 1 - 3 \left( -\frac{1}{4} \right) \right] = 2 - \frac{1}{4} \implies \frac{7}{4}A = \frac{7}{4} \implies A = 1 \end{align*}, we can then see
\displaystyle \displaystyle \begin{align*} \frac{2 + x}{(1 + 4x)(1 - 3x)} &= \frac{1}{1 + 4x} + \frac{1}{1 - 3x} \\ &= \frac{1}{1 - \left( -4x \right) } + \frac{1}{1 - 3x} \\ &= \sum_{m = 1}^{\infty} \left( -4x \right) ^m + \sum_{n = 1}^{\infty} \left( 3x \right) n \textrm{ if } |x| < \frac{1}{4} \textrm{ and } |x| < \frac{1}{3} \\ &= \sum_{n = 1}^{\infty} \left[ (-4x)^n + (3x)^n \right] \textrm{ if } |x| < \frac{1}{4} \end{align*}
So the first four terms will be \displaystyle \displaystyle \begin{align*} \left\{ -4x + 3x, (-4x)^2 + (3x)^2, (-4x)^3 + (3x)^3, (-4x)^4 + (3x)^4 \dots \right\} = \left\{ -x, 25x^2, -55x^3, 331x^4 \dots \right\} \end{align*}.