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Math Help - Rational expression and binomial expansion

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    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.
    Please help me on how to solve this
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    Re: Rational expression and binomial expansion

    Sorry guys, already solved it myself
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    Re: Rational expression and binomial expansion

    Quote Originally Posted by Bernana View Post
    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.
    Please help me on how to solve this
    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*}.
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