A sequence $u_1, u_2, u_3, \dots$ satisfies the recurrence relation $u_n = \frac{n}{(n-1)^2}u_{n-1}$, for $n \ge 2$.

1. State the statement and verify the base case for $n=1$.

Let $P_n$ be the statement that $u_n = \frac{2n}{(n-1)!}$ for integers $n \ge 1$.

For $n = 1$,
$\text{LHS} = u_1 = 2$ (given)
$\text{RHS} = \frac{2(1)}{(1-1)!} = \frac{2}{0!} = \frac{2}{1} = 2$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

2. State the inductive hypothesis.

Assume that $P_k$ is true for some positive integer $k \ge 1$.
That is, $u_k = \frac{2k}{(k-1)!}$.

3. Use the recurrence relation to show $P_{k+1}$ is true.

We want to show that $P_{k+1}$ is true, which states that $u_{k+1} = \frac{2(k+1)}{k!}$.
Using the given recurrence relation $u_n = \frac{n}{(n-1)^2}u_{n-1}$ for $n = k+1 \ge 2$:
$\begin{aligned} u_{k+1} &= \frac{k+1}{k^2} u_k \\[1ex] &= \frac{k+1}{k^2} \left[ \frac{2k}{(k-1)!} \right] \\[1ex] &= \frac{2k(k+1)}{k^2 (k-1)!} \\[1ex] &= \frac{2(k+1)}{k(k-1)!} \\[1ex] &= \frac{2(k+1)}{k!} \end{aligned}$
Thus, $P_{k+1}$ is true if $P_k$ is true.

4. State the standard induction conclusion.

Since $P_1$ is true, and $P_k$ is true implies that $P_{k+1}$ is true, it follows by the Principle of Mathematical Induction that $P_n$ is true for all integers $n \ge 1$.

1. Compute the values of $u_2$, $u_3$, and $u_4$ using the given recurrence relation.

Using the given recurrence relation $u_n = \frac{n}{(n-1)^2}u_{n-1}$ and $u_1 = a$:

For $n = 2$:
$u_2 = \frac{2}{1^2}u_1 = 2(a) = 2a$

For $n = 3$:
$u_3 = \frac{3}{2^2}u_2 = \frac{3}{4}(2a) = \frac{3}{2}a$

For $n = 4$:
$u_4 = \frac{4}{3^2}u_3 = \frac{4}{9}\left(\frac{3}{2}a\right) = \frac{2}{3}a$

2. Deduce the expression for $u_n$ in terms of $a$ by recognising the pattern or using the result from part (a).

We can rewrite the terms to reveal the pattern, guided by the result $u_n = \frac{2n}{(n-1)!}$ proven in part (a):
$u_1 = \frac{1 \cdot a}{1}$
$u_2 = \frac{2 \cdot a}{1}$
$u_3 = \frac{3 \cdot a}{2}$
$u_4 = \frac{4 \cdot a}{6}$

Notice that the denominators are $1, 1, 2, 6, \dots$ which corresponds to $(n-1)!$, while the numerators are $n \cdot a$.
Therefore,
$u_n = \frac{an}{(n-1)!}$

(Alternatively, since the recurrence relation $u_n = f(n)u_{n-1}$ implies that $u_n$ is directly proportional to the first term $u_1$, we can simply replace the initial value $2$ with $a$ in the formula derived in part (a).)