Let's solve part (ii) of the question. ii) Obtain the first four terms of the binomial expansion of $(1 - \frac{3}{2}x)^{-2}$ and find the set of values of $x$ for which the expansion is valid. The binomial expansion for $(1+y)^n$ is given by: $$(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots$$ In this case, $y = -\frac{3}{2}x$ and $n = -2$. Step 1: Find the first four terms of the expansion. The first term is $1$. The second term is $ny$: $$(-2)\left(-\frac{3}{2}x\right) = 3x$$ The third term is $\frac{n(n-1)}{2!}y^2$: $$\frac{(-2)(-2-1)}{2 \times 1}\left(-\frac{3}{2}x\right)^2 = \frac{(-2)(-3)}{2}\left(\frac{9}{4}x^2\right) = \frac{6}{2}\left(\frac{9}{4}x^2\right) = 3\left(\frac{9}{4}x^2\right) = \frac{27}{4}x^2$$ The fourth term is $\frac{n(n-1)(n-2)}{3!}y^3$: $$\frac{(-2)(-2-1)(-2-2)}{3 \times 2 \times 1}\left(-\frac{3}{2}x\right)^3 = \frac{(-2)(-3)(-4)}{6}\left(-\frac{27}{8}x^3\right) = \frac{-24}{6}\left(-\frac{27}{8}x^3\right) = -4\left(-\frac{27}{8}x^3\right) = \frac{108}{8}x^3 = \frac{27}{2}x^3$$ So, the first four terms of the expansion are: $$1 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3$$ Step 2: Find the set of values of $x$ for which the expansion is valid. The binomial expansion $(1+y)^n$ is valid when $|y| < 1$. In this expansion, $y = -\frac{3}{2}x$. So, we must have: $$\left|-\frac{3}{2}x\right| < 1$$ $$\frac{3}{2}|x| < 1$$ $$|x| < \frac{1}{3/2}$$ $$|x| < \frac{2}{3}$$ This inequality can also be written as: $$-\frac{2}{3} < x < \frac{2}{3}$$ The first four terms of the expansion are $\boxed{1 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3}$. The expansion is valid for $\boxed{|x| < \frac{2}{3}}$ or $\boxed{-\frac{2}{3} < x < \frac{2}{3}}$. What's next?