(a) Expand \((2 - 3x)^{-2}\) in ascending powers of \(x\), up to and including the term in \(x^2\), simplifying the coefficients.
(b) State the set of values of \(x\) for which the expansion is valid.
Solution
(a) To expand \((2 - 3x)^{-2}\), we can use the binomial series expansion for negative exponents:
\((1 - u)^{-n} = 1 + nu + \frac{n(n+1)}{2!}u^2 + \cdots\)
Here, \(u = \frac{3}{2}x\) and \(n = 2\).
First term: \(\frac{1}{4}\).
Second term: \(2 \times \frac{3}{2}x = 3x\), so the coefficient is \(\frac{3}{4}x\).
Third term: \(\frac{2 \times 3}{2} \times \left(\frac{3}{2}x\right)^2 = \frac{27}{16}x^2\).
Thus, the expansion up to \(x^2\) is \(\frac{1}{4} + \frac{3}{4}x + \frac{27}{16}x^2\).
(b) The expansion is valid for \(|\frac{3}{2}x| < 1\), which simplifies to \(|x| < \frac{2}{3}\).
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