已知对于任意非负整数 $n$，$u_0=\frac{1}{2}$ 且 $u_{n+1}=\dfrac{3u_n}{2u_n+1}$，

1. Evaluate the first four terms using the given recurrence relation.

Given $u_0 = \frac{1}{2}$ and $u_{n+1} = \frac{3u_n}{2u_n+1}$:

$u_1 = \frac{3(\frac{1}{2})}{2(\frac{1}{2})+1} = \frac{3/2}{2} = \frac{3}{4}$

$u_2 = \frac{3(\frac{3}{4})}{2(\frac{3}{4})+1} = \frac{9/4}{10/4} = \frac{9}{10}$

$u_3 = \frac{3(\frac{9}{10})}{2(\frac{9}{10})+1} = \frac{27/10}{28/10} = \frac{27}{28}$

$u_4 = \frac{3(\frac{27}{28})}{2(\frac{27}{28})+1} = \frac{81/28}{82/28} = \frac{81}{82}$

2. Express the evaluated terms in the form $\frac{f(n)}{1+f(n)}$ where $f(n)$ is explicitly in terms of $n$.

$u_1 = \frac{3}{1+3} = \frac{3^1}{1+3^1}$

$u_2 = \frac{9}{1+9} = \frac{3^2}{1+3^2}$

$u_3 = \frac{27}{1+27} = \frac{3^3}{1+3^3}$

$u_4 = \frac{81}{1+81} = \frac{3^4}{1+3^4}$

3. Formulate a general conjecture for $u_n$.

Observing that $f(n) = 3^n$, we can generalise the pattern for the $n$th term:
$u_n = \frac{3^n}{1+3^n}$

1. State the conjecture derived from part (i) and verify the base case for $n=0$.

From (i), the conjecture is $u_n = \dfrac{3^n}{1+3^n}$.
Let $P_n$ be the statement that $u_n = \dfrac{3^n}{1+3^n}$.

For $n=0$:
$\text{LHS} = u_0 = \frac{1}{2} \text{ (given)}$
$\text{RHS} = \frac{3^0}{1+3^0} = \frac{1}{1+1} = \frac{1}{2}$
Since $\text{LHS} = \text{RHS}$, $P_0$ is true.

2. State the inductive hypothesis.

Assume that $P_k$ is true for some non-negative integer $k$, i.e.,
$u_k = \frac{3^k}{1+3^k}$

3. Express $u_{k+1}$ using the given recurrence relation and substitute the inductive hypothesis.

We want to show that $P_{k+1}$ is true. Using the given recurrence relation:
$u_{k+1} = \frac{3u_k}{2u_k+1}$
Substituting our assumption for $u_k$:
$u_{k+1} = \frac{3\left(\dfrac{3^k}{1+3^k}\right)}{2\left(\dfrac{3^k}{1+3^k}\right)+1}$

4. Simplify the expression to establish $P_{k+1}$.

Multiply the numerator and the denominator by $(1+3^k)$:
$u_{k+1} = \frac{3(3^k)}{2(3^k) + (1+3^k)}$
$u_{k+1} = \frac{3^{k+1}}{3 \cdot 3^k + 1}$
$u_{k+1} = \frac{3^{k+1}}{1+3^{k+1}}$
This matches the formula for $n=k+1$. Hence, $P_k \implies P_{k+1}$.

5. Provide the final concluding statement for mathematical induction.

Since $P_0$ is true, and $P_k$ is true implies $P_{k+1}$ is true, by the principle of mathematical induction, $P_n$ is true for all non-negative integers $n$.