# Rational expression and binomial expansion

• Sep 28th 2013, 02:43 AM
Bernana
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.
• Sep 28th 2013, 02:59 AM
Bernana
Re: Rational expression and binomial expansion
Sorry guys, already solved it myself
• Sep 28th 2013, 04:11 AM
Prove It
Re: Rational expression and binomial expansion
Quote:

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.

Your approach seems fine. Attempting Partial Fractions:

\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 \begin{align*} x = \frac{1}{3} \end{align*} and we find \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 \begin{align*} x = -\frac{1}{4} \end{align*} and we find \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 \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 \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*}.