考虑 $S_{n}=\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\ldots+\dfrac{2}{n(n+1)(n+2)}$。

1. Calculate the first term of the series, $S_1$.

$S_1 = \dfrac{2}{1\times2\times3} = \dfrac{2}{6} = \dfrac{1}{3}$

2. Calculate the sum of the first two terms, $S_2$.

$S_2 = S_1 + \dfrac{2}{2\times3\times4} = \dfrac{1}{3} + \dfrac{2}{24} = \dfrac{1}{3} + \dfrac{1}{12} = \dfrac{5}{12}$

3. Calculate the sum of the first three terms, $S_3$.

$S_3 = S_2 + \dfrac{2}{3\times4\times5} = \dfrac{5}{12} + \dfrac{2}{60} = \dfrac{5}{12} + \dfrac{1}{30} = \dfrac{27}{60} = \dfrac{9}{20}$

1. Calculate the first three terms of the sequence as required from part (i).

$S_{1}=\frac{2}{1\times2\times3}=\frac{1}{3}$

$S_{2}=\frac{1}{3}+\frac{2}{2\times3\times4}=\frac{1}{3}+\frac{1}{12}=\frac{5}{12}$

$S_{3}=\frac{5}{12}+\frac{2}{3\times4\times5}=\frac{5}{12}+\frac{1}{30}=\frac{9}{20}$

2. Compare the terms with the given form to deduce the values of $a$ and $b$.

We are given the form $S_n = \frac{1}{2}-\frac{1}{(n+a)(n+b)}$. Rearranging gives $\frac{1}{(n+a)(n+b)} = \frac{1}{2} - S_n$.

For $n=1$: $\frac{1}{2} - \frac{1}{3} = \frac{1}{6} \implies (1+a)(1+b) = 6 = 2 \times 3$
For $n=2$: $\frac{1}{2} - \frac{5}{12} = \frac{1}{12} \implies (2+a)(2+b) = 12 = 3 \times 4$
For $n=3$: $\frac{1}{2} - \frac{9}{20} = \frac{1}{20} \implies (3+a)(3+b) = 20 = 4 \times 5$

By comparing the factors, we observe that the denominator follows the pattern $(n+1)(n+2)$. Hence $a=1$ and $b=2$ (or vice versa).

1. Determine the conjecture for $S_n$ from earlier parts.

From part (i), $S_1 = \frac{1}{3}$, $S_2 = \frac{5}{12}$, $S_3 = \frac{9}{20}$.
Equating these to $\frac{1}{2} - \frac{1}{(n+a)(n+b)}$:
For $n=1$, $\frac{1}{2} - \frac{1}{(1+a)(1+b)} = \frac{1}{3} \implies (1+a)(1+b) = 6$
For $n=2$, $\frac{1}{2} - \frac{1}{(2+a)(2+b)} = \frac{5}{12} \implies (2+a)(2+b) = 12$
For $n=3$, $\frac{1}{2} - \frac{1}{(3+a)(3+b)} = \frac{9}{20} \implies (3+a)(3+b) = 20$
By observation, $a=1$ and $b=2$ (or vice versa) satisfy all three equations.
Conjecture: $S_n = \frac{1}{2} - \frac{1}{(n+1)(n+2)}$.

2. Define the proposition and prove the base case $n=1$.

Let $P_n$ be the proposition that $S_n = \frac{1}{2} - \frac{1}{(n+1)(n+2)}$ for all $n \in \mathbb{Z}^+$.
When $n=1$:
$\text{LHS} = S_1 = \frac{1}{3}$ (from part i)
$\text{RHS} = \frac{1}{2} - \frac{1}{(1+1)(1+2)} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

3. State the induction hypothesis and express $S_{k+1}$.

Assume that $P_k$ is true for some $k \in \mathbb{Z}^+$, i.e.,
$S_k = \frac{1}{2} - \frac{1}{(k+1)(k+2)}$

For $n=k+1$, using the sequence definition $S_{k+1} = S_k + u_{k+1}$:
$S_{k+1} = \left[ \frac{1}{2} - \frac{1}{(k+1)(k+2)} \right] + \frac{2}{(k+1)(k+2)(k+3)}$

4. Perform the algebraic manipulation to show that $P_{k+1}$ is true, and write the concluding statement.

$S_{k+1} = \frac{1}{2} + \left[ \frac{-(k+3) + 2}{(k+1)(k+2)(k+3)} \right]$
$S_{k+1} = \frac{1}{2} + \left[ \frac{-k-1}{(k+1)(k+2)(k+3)} \right]$
$S_{k+1} = \frac{1}{2} - \frac{k+1}{(k+1)(k+2)(k+3)}$
$S_{k+1} = \frac{1}{2} - \frac{1}{(k+2)(k+3)}$
$S_{k+1} = \frac{1}{2} - \frac{1}{((k+1)+1)((k+1)+2)}$
Thus, $P_{k+1}$ is true if $P_k$ is true.

Since $P_1$ is true, and $P_k \implies P_{k+1}$, by the Principle of Mathematical Induction, $P_n$ is true for all $n \in \mathbb{Z}^+$.