Using the method of differences, find $\displaystyle \sum_{r=2}^{n} \dfrac{2}{r^{2}-1}$. Hence, find $\displaystyle \sum_{r=11}^{3n-1} \dfrac{2}{r(r+2)}$.

1. Express the general term in partial fractions.

$\frac{2}{r^2-1} = \frac{2}{(r-1)(r+1)}$
Let $\frac{2}{(r-1)(r+1)} \equiv \frac{A}{r-1} + \frac{B}{r+1}$
$2 = A(r+1) + B(r-1)$
Substitute $r = 1$: $2 = 2A \implies A = 1$
Substitute $r = -1$: $2 = -2B \implies B = -1$
Therefore, $\frac{2}{r^2-1} = \frac{1}{r-1} - \frac{1}{r+1}$

2. Write out the first few and last few terms of the series to identify the cancellation pattern.

$\sum_{r=2}^{n} \left( \frac{1}{r-1} - \frac{1}{r+1} \right)$
$= \left( \frac{1}{1} - \frac{1}{3} \right) \quad \text{(when } r=2\text{)}$
$+ \left( \frac{1}{2} - \frac{1}{4} \right) \quad \text{(when } r=3\text{)}$
$+ \left( \frac{1}{3} - \frac{1}{5} \right) \quad \text{(when } r=4\text{)}$
$+ \dots$
$+ \left( \frac{1}{n-2} - \frac{1}{n} \right) \quad \text{(when } r=n-1\text{)}$
$+ \left( \frac{1}{n-1} - \frac{1}{n+1} \right) \quad \text{(when } r=n\text{)}$

3. Cancel out the terms and simplify to obtain the final sum.

Observing the pattern, terms cancel out with a gap of two. The first two positive terms and the last two negative terms survive the cancellation.
$= \frac{1}{1} + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1}$
$= \frac{3}{2} - \frac{1}{n} - \frac{1}{n+1}$

1. Determine the result of part (a) since it must be used to solve the target

Using the method of differences for part (a):
$\dfrac{2}{r^2-1} = \dfrac{1}{r-1} - \dfrac{1}{r+1}$
$\sum_{r=2}^{n} \dfrac{2}{r^{2}-1} = \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n-1} - \frac{1}{n+1} \right)$
$= 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1} = \frac{3}{2} - \frac{1}{n} - \frac{1}{n+1}$

2. Relate the target series to the series in part (a) by substituting the index

Let $k = r + 1$, which means $r = k - 1$.
When $r = 11$, $k = 12$.
When $r = 3n-1$, $k = 3n$.
Substituting these into the sum:
$\sum_{r=11}^{3n-1} \dfrac{2}{r(r+2)} = \sum_{k=12}^{3n} \dfrac{2}{(k-1)(k+1)} = \sum_{k=12}^{3n} \dfrac{2}{k^2-1}$

3. Express the new sum as a difference of two sums starting from the original lower bound of $k=2$

$\sum_{k=12}^{3n} \dfrac{2}{k^2-1} = \sum_{k=2}^{3n} \dfrac{2}{k^2-1} - \sum_{k=2}^{11} \dfrac{2}{k^2-1}$

4. Apply the result from part (a) to both sums and simplify

Using the formula $\sum_{r=2}^{N} \frac{2}{r^{2}-1} = \frac{3}{2} - \frac{1}{N} - \frac{1}{N+1}$ for $N=3n$ and $N=11$:
$\sum_{k=12}^{3n} \dfrac{2}{k^2-1} = \left( \frac{3}{2} - \frac{1}{3n} - \frac{1}{3n+1} \right) - \left( \frac{3}{2} - \frac{1}{11} - \frac{1}{12} \right)$
$= \frac{1}{11} + \frac{1}{12} - \frac{1}{3n} - \frac{1}{3n+1}$
$= \frac{23}{132} - \frac{1}{3n} - \frac{1}{3n+1}$