Write down the first four terms in the expansion of $\left(1 - 4x\right)^{6}$. Hence find the coefficient of $x^{3}$ in the expansion of $\left(3 + x^{2}\right)\left(1 - 4x\right)^{6}$. Explain why there is no constant term in the expansion of $\left(2x + \frac{1}{x^{3}}\right)^{18}$.

1. Apply the binomial theorem to expand the first four terms

$\left(1 - 4x\right)^{6} = 1 + \binom{6}{1}(-4x) + \binom{6}{2}(-4x)^2 + \binom{6}{3}(-4x)^3 + \dots$

2. Evaluate the binomial coefficients and simplify the terms

$= 1 + 6(-4x) + 15(16x^2) + 20(-64x^3) + \dots$
$= 1 - 24x + 240x^2 - 1280x^3$

1. Recall or calculate the first four terms of the expansion of $(1 - 4x)^6$ from part (a)(i)

From part (a)(i),
$(1 - 4x)^6 = 1 + \binom{6}{1}(-4x) + \binom{6}{2}(-4x)^2 + \binom{6}{3}(-4x)^3 + \dots$
$= 1 - 24x + 240x^2 - 1280x^3 + \dots$

2. Identify the products that result in an $x^3$ term when multiplying $(3 + x^2)$ by the expansion

We want the coefficient of $x^3$ in:
$(3 + x^2)(1 - 24x + 240x^2 - 1280x^3 + \dots)$
The $x^3$ terms are formed by:
$3 \times (-1280x^3) = -3840x^3$
$x^2 \times (-24x) = -24x^3$

3. Calculate the final coefficient of $x^3$

Coefficient of $x^3 = -3840 - 24 = -3864$

1. Write down the general term of the expansion

The general term is given by:
$T_{r+1} = \binom{18}{r} (2x)^{18-r} \left(\frac{1}{x^3}\right)^r$
$T_{r+1} = \binom{18}{r} 2^{18-r} x^{18-r} x^{-3r}$
$T_{r+1} = \binom{18}{r} 2^{18-r} x^{18-4r}$

2. Set the power of $x$ to 0 to find the condition for a constant term

For a constant term, the power of $x$ must be 0:
$18 - 4r = 0$
$4r = 18$
$r = 4.5$

3. Explain why the result means there is no constant term

Since $r$ must be an integer ($0 \leqslant r \leqslant 18$) and $4.5$ is not an integer, there is no value of $r$ that yields a constant term. Therefore, there is no constant term in the expansion.