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I need to find the power series expansion of $$(2-3x)^{-1/3}$$ By using the Generalized Binomial Theorem: $$(1+x)^m= \sum_{n\geq0}^{n}\dbinom{m}{n}x^n$$

I am told to use the following definition of binomial coefficients where m and n are real numbers: $$\dbinom{m}{n} = \frac{m(m-1)..(m-n+1)}{n!}$$

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    $\begingroup$ Have you tried anything? $\endgroup$ – Luke Collins Feb 22 '20 at 1:17
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Just use the formula... \begin{align*} (2-3x)^{-1/3} &= 2^{-1/3}\big(1 + (-\tfrac{3x}{2})\big)^{-1/3}\\ &= \frac{1}{\sqrt[3]2}\bigg(1 + (-\tfrac13)(-\tfrac{3x}{2}) + \frac{(-\frac13)(-\frac43)}{2!}(-\tfrac{3x}{2})^2 + \frac{(-\frac13)(-\frac43)(-\frac73)}{3!}(-\tfrac{3x}{2})^3+\cdots \bigg)\\ &= \frac{1}{\sqrt[3]2}\bigg(1 + \frac x2 + \frac{x^2}2 + \frac{7x^3}{12} \bigg) + O(x^4), \end{align*} valid for $|-\frac{3x}{2}|<1$, i.e., for $|x|<\frac23$.


Note: If you are unfamiliar with the big-Oh notation (${}+O(x^4)$ above), this simply represents an "error" no bigger than some constant times $x^4$ (recall that for $x$ in the range of convergence, higher powers are smaller). If this is confusing, just substitute the more hand-wavy "${}+\cdots$" instead.

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