In the expansion of $\left(x^2 - \frac{1}{2x^4}\right)^n$, the sixth term is independent of $x$.

1. Write down the expression for the sixth term

The general term is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
For the sixth term, $r = 5$.
$T_6 = \binom{n}{5} (x^2)^{n-5} \left(-\frac{1}{2x^4}\right)^5$

2. Simplify the expression to isolate the power of $x$

$T_6 = \binom{n}{5} x^{2n-10} \left(-\frac{1}{32}\right) x^{-20}$
$T_6 = \binom{n}{5} \left(-\frac{1}{32}\right) x^{2n-30}$

3. Equate the power of $x$ to zero to show $n=15$

Since the term is independent of $x$, the power of $x$ must be $0$.
$2n - 30 = 0$
$2n = 30$
$n = 15 \text{ (shown)}$

4. Calculate the value of the sixth term

Substitute $n = 15$ into the coefficient:
$T_6 = \binom{15}{5} \left(-\frac{1}{32}\right)$
$T_6 = 3003 \times \left(-\frac{1}{32}\right) = -\frac{3003}{32}$

1. Identify the terms required from the expansion of $\left(x^2 - \frac{1}{2x^4}\right)^{15}$

To find the term independent of $x$ in $\left(x^2 - \frac{1}{2x^4}\right)^{15} (32 + 64x^6)$, we need to multiply the constant term from the binomial expansion by $32$, and the $x^{-6}$ term from the binomial expansion by $64x^6$.

2. Find the constant term in the binomial expansion (from part i)

From part (i), the term independent of $x$ occurs when $r=5$:
$T_6 = \binom{15}{5} (x^2)^{10} \left(-\frac{1}{2x^4}\right)^5 = 3003 \left(-\frac{1}{32}\right) = -\frac{3003}{32}$

3. Find the term in $x^{-6}$ in the binomial expansion

Using the general term $T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(-\frac{1}{2x^4}\right)^r$, the power of $x$ is $2(15-r) - 4r = 30 - 6r$.
For the $x^{-6}$ term, set $30 - 6r = -6 \implies 6r = 36 \implies r = 6$.
The term is $T_7 = \binom{15}{6} (x^2)^9 \left(-\frac{1}{2x^4}\right)^6 = 5005 \left(\frac{1}{64}\right) x^{-6} = \frac{5005}{64} x^{-6}$.

4. Calculate the final term independent of $x$

Term independent of $x$ = $32 \left(-\frac{3003}{32}\right) + 64 \left(\frac{5005}{64}\right)$
$= -3003 + 5005$
$= 2002$