Write down the general term, $T_{r+1}$, for the binomial expansion of $\left(2x + \dfrac{1}{2x^3}\right)^{12}$. Find the term independent of $x$ in the expansion of $\left(\dfrac{1}{x^4} - 1\right)\left(2x + \dfrac{1}{2x^3}\right)^{12}$. Explain whether there are any terms containing $x^n$, where $n$ is odd in the expansion of $\left(2x + \dfrac{1}{2x^3}\right)^{12}$.

1. Write down the unsimplified general term using the binomial theorem

$T_{r+1} = \binom{12}{r} (2x)^{12-r} \left(\frac{1}{2x^3}\right)^r$

2. Simplify the general term by grouping the constants and the powers of $x$

$T_{r+1} = \binom{12}{r} 2^{12-r} x^{12-r} \left(2^{-1}\right)^r x^{-3r} \\ = \binom{12}{r} 2^{12-2r} x^{12-4r}$

1. Write down the general term of the binomial expansion from part (a)

The general term of $\left(2x + \dfrac{1}{2x^3}\right)^{12}$ is:
$T_{r+1} = \binom{12}{r} (2x)^{12-r} \left(\dfrac{1}{2x^3}\right)^r$
$T_{r+1} = \binom{12}{r} 2^{12-r} x^{12-r} \cdot 2^{-r} x^{-3r}$
$T_{r+1} = \binom{12}{r} 2^{12-2r} x^{12-4r}$

2. Identify the powers of $x$ needed from the expansion

To find the term independent of $x$ in $\left(\dfrac{1}{x^4} - 1\right)\left(2x + \dfrac{1}{2x^3}\right)^{12}$, we need to multiply $\dfrac{1}{x^4}$ by the term in $x^4$, and $-1$ by the term in $x^0$ (the constant term).

3. Find the term in $x^4$

For the term in $x^4$:
$12 - 4r = 4$
$4r = 8 \implies r = 2$
The term is:
$T_3 = \binom{12}{2} 2^{12-4} x^4 = 66 \times 2^8 x^4 = 16896x^4$

4. Find the term in $x^0$

For the term in $x^0$:
$12 - 4r = 0$
$4r = 12 \implies r = 3$
The term is:
$T_4 = \binom{12}{3} 2^{12-6} x^0 = 220 \times 2^6 = 14080$

5. Calculate the final term independent of $x$

The term independent of $x$ is:
$\left(\dfrac{1}{x^4}\right)(16896x^4) + (-1)(14080)$
$= 16896 - 14080$
$= 2816$

1. Recall or derive the power of x in the general term from part (a)

From part (a), the general term is $T_{r+1} = \binom{12}{r} (2x)^{12-r} \left(\frac{1}{2x^3}\right)^r$.
The power of $x$ is $(12 - r) - 3r = 12 - 4r$.

2. Explain why the power of x cannot be odd

The power of $x$ is $12 - 4r = 4(3 - r)$.
Since $r$ is an integer ($0 \leqslant r \leqslant 12$), $4(3 - r)$ is a multiple of 4, which is always an even number.
Therefore, $12 - 4r$ cannot be an odd number $n$.