By writing $\frac{3}{(3r+1)(3r+4)}$ in partial fractions, find $\sum_{r=1}^{n} \frac{3}{(3r+1)(3r+4)}$ in terms of $n$. Hence find the exact value of $\frac{3}{22} \times \frac{1}{25} + \frac{3}{25} \times \frac{1}{28} + \frac{3}{28} \times \frac{1}{31} + \cdots$.

1. Express the general term in partial fractions.

Let $\frac{3}{(3r+1)(3r+4)} = \frac{A}{3r+1} + \frac{B}{3r+4}$

$3 = A(3r+4) + B(3r+1)$

Substituting $r = -\frac{1}{3}$, we get $3 = A\left(3(-\frac{1}{3}) + 4\right) \implies 3 = 3A \implies A = 1$

Substituting $r = -\frac{4}{3}$, we get $3 = B\left(3(-\frac{4}{3}) + 1\right) \implies 3 = -3B \implies B = -1$

Therefore, $\frac{3}{(3r+1)(3r+4)} = \frac{1}{3r+1} - \frac{1}{3r+4}$

2. List out the first few terms and the last few terms of the sum to identify the cancellation pattern.

$\sum_{r=1}^{n} \frac{3}{(3r+1)(3r+4)} = \sum_{r=1}^{n} \left( \frac{1}{3r+1} - \frac{1}{3r+4} \right)$

$= \left( \frac{1}{4} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{10} \right) + \left( \frac{1}{10} - \frac{1}{13} \right) + \dots$
$\dots + \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right) + \left( \frac{1}{3n+1} - \frac{1}{3n+4} \right)$

3. Cancel the intermediate terms to obtain the simplified sum.

Observing that adjacent terms cancel out (e.g. $-\frac{1}{7}$ and $+\frac{1}{7}$), only the very first and the very last terms remain:

$= \frac{1}{4} - \frac{1}{3n+4}$

1. Establish the partial sums formula from part (i) as it is necessary to solve this 'Hence' part.

By resolving into partial fractions:
$\frac{3}{(3r+1)(3r+4)} = \frac{1}{3r+1} - \frac{1}{3r+4}$
Using the method of differences:
$\sum_{r=1}^{n} \frac{3}{(3r+1)(3r+4)} = \sum_{r=1}^{n} \left( \frac{1}{3r+1} - \frac{1}{3r+4} \right)$
$= \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \dots + \left(\frac{1}{3n+1} - \frac{1}{3n+4}\right)$
$= \frac{1}{4} - \frac{1}{3n+4}$

2. Identify the starting term of the given series and find the sum to infinity of the original series from r = 1.

The given series is $\frac{3}{22 \times 25} + \frac{3}{25 \times 28} + \dots$
Equating the first factor to the general term, $3r+1 = 22$, yields $r = 7$.
Using the result from (i), the sum to infinity ($S_\infty$) of the entire series is:
$S_\infty = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{3}{(3r+1)(3r+4)} = \lim_{n \to \infty} \left( \frac{1}{4} - \frac{1}{3n+4} \right) = \frac{1}{4}$

3. Calculate the exact value of the required series by subtracting the sum of the first 6 terms from the sum to infinity.

The sum from $r=7$ to $\infty$ can be evaluated by computing $S_\infty - S_6$.
Using the formula from (i) for $n=6$:
$S_6 = \frac{1}{4} - \frac{1}{3(6)+4} = \frac{1}{4} - \frac{1}{22}$
Therefore, the exact value of the required series is:
$\sum_{r=7}^{\infty} \frac{3}{(3r+1)(3r+4)} = S_\infty - S_6 = \frac{1}{4} - \left( \frac{1}{4} - \frac{1}{22} \right) = \frac{1}{22}$