Explain why all terms in the expansion of $\left(kx^3 + x\right)^{12}$ do not contain any odd powers of $x$. Given that the coefficient of $x^{16}$ in the expansion of $\left(1 - \frac{x}{2}\right)^2 \left(kx^3 + x\right)^{12}$ is $258$, find the integral value of $k$.

1. Write down the general term of the binomial expansion.

The general term is given by:
$T_{r+1} = \binom{12}{r} (kx^3)^{12-r} (x)^r$

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

$T_{r+1} = \binom{12}{r} k^{12-r} x^{3(12-r)} x^r$
$T_{r+1} = \binom{12}{r} k^{12-r} x^{36-3r+r}$
$T_{r+1} = \binom{12}{r} k^{12-r} x^{36-2r}$

3. Explain why the power of $x$ is always even.

The power of $x$ is $36 - 2r = 2(18 - r)$.
Since $r$ is an integer ($0 \leqslant r \leqslant 12$), $18 - r$ is an integer, making $2(18 - r)$ an even integer.
Therefore, all terms have an even power of $x$, meaning the expansion does not contain any odd powers of $x$.

1. Expand the first bracket $\left(1 - \frac{x}{2}\right)^2$

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

2. Write down the general term for the expansion of $(kx^3 + x)^{12}$

$T_{r+1} = \binom{12}{r} (kx^3)^{12-r} (x)^r = \binom{12}{r} k^{12-r} x^{36-3r+r} = \binom{12}{r} k^{12-r} x^{36-2r}$

3. Identify the terms from $(kx^3 + x)^{12}$ needed to form $x^{16}$ when multiplied by $\left(1 - x + \frac{x^2}{4}\right)$

The terms in the first bracket have powers $x^0$, $x^1$, and $x^2$.
To get $x^{16}$, we need terms from $(kx^3 + x)^{12}$ with powers $x^{16}$, $x^{15}$, and $x^{14}$.
From part (a), all powers of $x$ in $(kx^3 + x)^{12}$ are even, so there is no $x^{15}$ term.
We need $36 - 2r = 16 \implies 2r = 20 \implies r = 10$.
We need $36 - 2r = 14 \implies 2r = 22 \implies r = 11$.

4. Calculate the coefficients of $x^{16}$ and $x^{14}$ in $(kx^3 + x)^{12}$

For $r=10$ (power 16): Coefficient is $\binom{12}{10} k^{12-10} = 66k^2$.
For $r=11$ (power 14): Coefficient is $\binom{12}{11} k^{12-11} = 12k$.

5. Form an equation for the coefficient of $x^{16}$ in the full expansion and solve for $k$

Coefficient of $x^{16}$ in the product is:
$1 \times (66k^2) + \frac{1}{4} \times (12k) = 258$
$66k^2 + 3k - 258 = 0$
$22k^2 + k - 86 = 0$
$(22k + 43)(k - 2) = 0$
Since $k$ is an integer, $k = 2$.