Write down and simplify the first three terms in the expansion, in ascending powers of $x$, of $\left(2 + \frac{x}{2}\right)^n$, where $n$ is a positive integer. The first two terms in the expansion, in ascending powers of $x$, of $\left(\frac{4}{3} - 3x\right)\left(2 + \frac{x}{2}\right)^n$ are $a + bx^2$. Find the value of $n$. Find the term independent of $x$ in the expansion of $\left(x - \frac{1}{2x^2}\right)^9$.

1. Apply the binomial theorem to expand the expression up to the $x^2$ term.

$\left(2 + \frac{x}{2}\right)^n = 2^n + \binom{n}{1}2^{n-1}\left(\frac{x}{2}\right) + \binom{n}{2}2^{n-2}\left(\frac{x}{2}\right)^2 + \dots$

2. Simplify the coefficients of the first three terms.

$= 2^n + n2^{n-1}\left(\frac{x}{2}\right) + \frac{n(n-1)}{2}2^{n-2}\left(\frac{x^2}{4}\right)$
$= 2^n + n2^{n-2}x + n(n-1)2^{n-5}x^2$

1. Write down the general term of the expansion

The general term $T_{r+1}$ in the expansion of $\left(x - \frac{1}{2x^2}\right)^9$ is given by:
$T_{r+1} = \binom{9}{r} (x)^{9-r} \left(-\frac{1}{2x^2}\right)^r$

2. Simplify the general term to isolate the power of $x$

$T_{r+1} = \binom{9}{r} x^{9-r} \left(-\frac{1}{2}\right)^r x^{-2r}$
$T_{r+1} = \binom{9}{r} \left(-\frac{1}{2}\right)^r x^{9-3r}$

3. Find the value of $r$ for the term independent of $x$

For the term independent of $x$, the power of $x$ must be $0$:
$9 - 3r = 0$
$3r = 9$
$r = 3$

4. Calculate the term independent of $x$

Substitute $r = 3$ into the coefficient part of the general term:
$T_4 = \binom{9}{3} \left(-\frac{1}{2}\right)^3$
$T_4 = 84 \times \left(-\frac{1}{8}\right)$
$T_4 = -10.5$