Express $\dfrac{5 + x^2}{(2 + x)(1 - x)^2}$ in the form of $\dfrac{A}{(2 + x)} + \dfrac{B}{(1 - x)^2}$, where $A$ and $B$ are constants to be found. Hence, expand $\dfrac{5 + x^2}{(2 + x)(1 - x)^2}$ as a series of ascending powers of $x$ up to and including the $x^2$ term. State the range of values of $x$ for which the expansion is valid.

1. Set up the algebraic identity by clearing the denominators.

$\frac{5 + x^2}{(2 + x)(1 - x)^2} = \frac{A}{2 + x} + \frac{B}{(1 - x)^2}$
Multiplying both sides by $(2 + x)(1 - x)^2$ yields:
$5 + x^2 = A(1 - x)^2 + B(2 + x)$

2. Substitute suitable values of $x$ to solve for $A$ and $B$.

Substitute $x = 1$ to eliminate $A$:
$5 + (1)^2 = A(0)^2 + B(2 + 1)$
$6 = 3B \implies B = 2$

Substitute $x = -2$ to eliminate $B$:
$5 + (-2)^2 = A(1 - (-2))^2 + B(0)$
$9 = 9A \implies A = 1$

3. State the final expression.

Substituting $A = 1$ and $B = 2$ back into the original form:
$\frac{1}{2 + x} + \frac{2}{(1 - x)^2}$

1. Determine the partial fractions from part (i) to use for the expansion.

From part (i), we solve $A(1 - x)^2 + B(2 + x) = 5 + x^2$.
Substituting $x = 1$, we get $3B = 6 \implies B = 2$.
Substituting $x = -2$, we get $9A = 9 \implies A = 1$.
Thus, $\dfrac{5 + x^2}{(2 + x)(1 - x)^2} = \dfrac{1}{2 + x} + \dfrac{2}{(1 - x)^2}$.

2. Expand the first term $\frac{1}{2+x}$ as a binomial series up to the $x^2$ term.

$\dfrac{1}{2+x} = (2+x)^{-1} = 2^{-1}\left(1+\dfrac{x}{2}\right)^{-1}$
$= \dfrac{1}{2} \left[ 1 + (-1)\left(\dfrac{x}{2}\right) + \dfrac{(-1)(-2)}{2!} \left(\dfrac{x}{2}\right)^2 + \dots \right]$
$= \dfrac{1}{2} \left( 1 - \dfrac{x}{2} + \dfrac{x^2}{4} \right)$
$= \dfrac{1}{2} - \dfrac{x}{4} + \dfrac{x^2}{8}$

3. Expand the second term $\frac{2}{(1-x)^2}$ as a binomial series up to the $x^2$ term.

$\dfrac{2}{(1-x)^2} = 2(1-x)^{-2}$
$= 2 \left[ 1 + (-2)(-x) + \dfrac{(-2)(-3)}{2!}(-x)^2 + \dots \right]$
$= 2 (1 + 2x + 3x^2)$
$= 2 + 4x + 6x^2$

4. Sum the two series expansions and simplify to find the final series.

$\dfrac{5 + x^2}{(2 + x)(1 - x)^2} \approx \left(\dfrac{1}{2} - \dfrac{x}{4} + \dfrac{x^2}{8}\right) + (2 + 4x + 6x^2)$
$= \left(\dfrac{1}{2} + 2\right) + \left(-\dfrac{1}{4} + 4\right)x + \left(\dfrac{1}{8} + 6\right)x^2$
$= \dfrac{5}{2} + \dfrac{15}{4}x + \dfrac{49}{8}x^2$

1. Identify the intersection of the validity ranges for both expanded terms.

From the partial fractions in part (i), the expression is $\frac{1}{2+x} + \frac{2}{(1-x)^2}$.
The expansion of $(2+x)^{-1} = \frac{1}{2}(1+\frac{x}{2})^{-1}$ is valid for $\left|\frac{x}{2}\right| < 1 \implies |x| < 2$.
The expansion of $(1-x)^{-2}$ is valid for $|-x| < 1 \implies |x| < 1$.
The expansion of the entire expression is valid for the intersection of these sets, which is $|x| < 1$.