It is given that $\displaystyle \sum_{r=1}^n \frac{2}{r(r+1)(r+2)} = \frac{1}{2} - \frac{1}{(n+1)(n+2)}$.

1. Evaluate the limit of the sum to $n$ terms as $n$ approaches infinity.

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1}{2} - \frac{1}{(n+1)(n+2)} \right)$

As $n \to \infty$, $\dfrac{1}{(n+1)(n+2)} \to 0$.

$\therefore S_\infty = \frac{1}{2}$

1. Express the required sum as the difference between two sums of the original series.

The series starts at $r=17$. The $n$-th term of this new series corresponds to $r=16+n$.
Let the required sum be $S'$.
$S' = \sum_{r=17}^{n+16} \frac{2}{r(r+1)(r+2)}$
$S' = \sum_{r=1}^{n+16} \frac{2}{r(r+1)(r+2)} - \sum_{r=1}^{16} \frac{2}{r(r+1)(r+2)}$

2. Substitute the appropriate limits into the given sum formula and simplify.

Using the given formula $\sum_{r=1}^n \frac{2}{r(r+1)(r+2)} = \frac{1}{2} - \frac{1}{(n+1)(n+2)}$:
$S' = \left( \frac{1}{2} - \frac{1}{(n+16+1)(n+16+2)} \right) - \left( \frac{1}{2} - \frac{1}{(16+1)(16+2)} \right)$
$S' = \left( \frac{1}{2} - \frac{1}{(n+17)(n+18)} \right) - \left( \frac{1}{2} - \frac{1}{17 \times 18} \right)$
$S' = \frac{1}{17 \times 18} - \frac{1}{(n+17)(n+18)}$
$S' = \frac{1}{306} - \frac{1}{(n+17)(n+18)}$

1. Factorise the denominator of the general term to relate it to the given identity.

$\sum_{r=2}^n \frac{1}{r(r^2-1)} = \sum_{r=2}^n \frac{1}{(r-1)r(r+1)}$

2. Adjust the index and introduce a constant factor to perfectly match the terms of the given series.

Let $k = r-1$. As $r$ goes from $2$ to $n$, $k$ goes from $1$ to $n-1$.
$\sum_{r=2}^n \frac{1}{(r-1)r(r+1)} = \frac{1}{2} \sum_{r=2}^n \frac{2}{(r-1)r(r+1)}$
$= \frac{1}{2} \sum_{k=1}^{n-1} \frac{2}{k(k+1)(k+2)}$

3. Apply the given sum formula by replacing $n$ with $n-1$.

Using the given identity $\displaystyle \sum_{r=1}^m \frac{2}{r(r+1)(r+2)} = \frac{1}{2} - \frac{1}{(m+1)(m+2)}$ with $m = n-1$:
$= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{(n-1+1)(n-1+2)} \right]$
$= \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{n(n+1)} \right]$
$= \frac{1}{4} - \frac{1}{2n(n+1)}$

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

$u_2 = 2 - \frac{4}{u_1} = 2 - \frac{4}{3} = \frac{2}{3}$
$u_3 = 2 - \frac{4}{u_2} = 2 - \frac{4}{\frac{2}{3}} = 2 - 6 = -4$
$u_4 = 2 - \frac{4}{u_3} = 2 - \frac{4}{-4} = 2 + 1 = 3$

2. Deduce the period of the sequence and evaluate $u_{100}$

Since $u_4 = u_1 = 3$, the sequence is periodic with a repeating block of 3 terms: $3, \frac{2}{3}, -4$.
To find $u_{100}$, divide 100 by the period length 3:
$100 = 33 \times 3 + 1$
The remainder is 1, so the 100th term is the 1st term in the repeating block.
$u_{100} = u_1 = 3$

1. Determine the first few terms (from part (b)(i)) to establish the periodicity of the sequence.

$u_1 = 3$
$u_2 = 2 - \frac{4}{3} = \frac{2}{3}$
$u_3 = 2 - \frac{4}{2/3} = 2 - 6 = -4$
$u_4 = 2 - \frac{4}{-4} = 3$
The sequence repeats every 3 terms.

2. Calculate the sum of one complete cycle of 3 terms.

$u_1 + u_2 + u_3 = 3 + \frac{2}{3} - 4 = -\frac{1}{3}$

3. Express the series sum using the number of full cycles and the remainder.

The number of terms is $3n-2$, which can be written as $3(n-1) + 1$.
This corresponds to $(n-1)$ full cycles of 3 terms, plus $1$ additional term (which is $u_1$).
$\sum_{r=1}^{3n-2} u_r = (n-1)(u_1 + u_2 + u_3) + u_1$
$= (n-1)\left(-\frac{1}{3}\right) + 3$
$= -\frac{1}{3}n + \frac{1}{3} + 3$
$= \frac{10 - n}{3}$