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Verify a summation formula for (12 – 5x) / (6 – 5x -x2)

Assume that \frac{12-5x}{6-5x-x^2} has a power-series representation in terms of powers of x, verify that it has the form

    \[ \frac{12-5x}{6-5x-x^2} = \sum_{n=0}^{\infty} \left( 1 + \frac{(-1)^n}{6^n} \right) x^n \]

valid for all real x such that |x| < 1.


We use partial fractions to decompose the given expression,

    \[ \frac{12-5x}{6-5x-x^2} = \frac{12-5x}{(1-x)(6+x)} = \frac{1}{1-x} + \frac{6}{6+x} = \frac{1}{1-x} + \frac{1}{1+\frac{x}{6}}. \]

These two terms are the sums of geometric series,

    \begin{align*}  \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n & \text{valid for } |x| < 1 \\  \frac{6}{6+x} &= \sum_{n=0}^{\infty} (-1)^n \left( \frac{x}{6} \right)^n & \text{valid for } |x| < 6. \end{align*}

Therefore, for all |x| < 1 we have

    \begin{align*}  \frac{12-5x}{6-5x-x^2} &= \frac{1}{1-x} + \frac{1}{1+\frac{x}{6}} \\[9pt]  &= \sum_{n=0}^{\infty} x^n + \sum_{n=0}^{\infty} (-1)^n \left( \frac{x}{6} \right)^n \\[9pt]  &= \sum_{n=0}^{\infty} \left( 1 + \left( \frac{(-1)}{6} \right)^n \right) x^n. \end{align*}

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