It is given that $\mathrm{f}(r) = 1 - \frac{3}{(r+2)!}$, where $r$ is a non-negative integer. Show that $\mathrm{f}(r) - \mathrm{f}(r-1) = \frac{a(r+1)}{(r+2)!}$ for some constant $a$ to be determined.

1. Determine the constant $a$ from the preliminary part and express the summand as a difference.

Since the previous part asks to determine $a$, we first evaluate $\mathrm{f}(r) - \mathrm{f}(r-1)$:
$\mathrm{f}(r) - \mathrm{f}(r-1) = \left( 1 - \frac{3}{(r+2)!} \right) - \left( 1 - \frac{3}{(r+1)!} \right)$
$= \frac{3}{(r+1)!} - \frac{3}{(r+2)!} = \frac{3(r+2) - 3}{(r+2)!} = \frac{3(r+1)}{(r+2)!}$
Thus, $a = 3$, which gives $\frac{r+1}{(r+2)!} = \frac{1}{3}[\mathrm{f}(r) - \mathrm{f}(r-1)]$.

2. Set up the summation and apply the method of differences.

$\sum_{r=1}^{n} \frac{r+1}{(r+2)!} = \frac{1}{3} \sum_{r=1}^{n} [\mathrm{f}(r) - \mathrm{f}(r-1)]$
$= \frac{1}{3} \Big[ (\mathrm{f}(1) - \mathrm{f}(0)) + (\mathrm{f}(2) - \mathrm{f}(1)) + \dots + (\mathrm{f}(n) - \mathrm{f}(n-1)) \Big]$

3. Cancel intermediate terms and substitute the function values.

Cancelling the intermediate terms leaves only the first and last terms:
$= \frac{1}{3} [\mathrm{f}(n) - \mathrm{f}(0)]$
Substituting $\mathrm{f}(n) = 1 - \frac{3}{(n+2)!}$ and $\mathrm{f}(0) = 1 - \frac{3}{2!} = -\frac{1}{2}$:
$= \frac{1}{3} \left[ \left( 1 - \frac{3}{(n+2)!} \right) - \left(-\frac{1}{2}\right) \right]$

4. Simplify the final algebraic expression.

$= \frac{1}{3} \left[ \frac{3}{2} - \frac{3}{(n+2)!} \right]$
$= \frac{1}{2} - \frac{1}{(n+2)!}$

1. Concisely calculate the result of part (i) as it is required to deduce the final answer.

From the main question stem, we first find $a$:
$\mathrm{f}(r) - \mathrm{f}(r-1) = \left( 1 - \frac{3}{(r+2)!} \right) - \left( 1 - \frac{3}{(r+1)!} \right)$
$= \frac{3}{(r+1)!} - \frac{3}{(r+2)!} = \frac{3(r+2) - 3}{(r+2)!} = \frac{3(r+1)}{(r+2)!}$
Thus, $a = 3$. For part (i), using the method of differences:
$\sum_{r=1}^{n} \frac{r+1}{(r+2)!} = \frac{1}{3} \sum_{r=1}^{n} [\mathrm{f}(r) - \mathrm{f}(r-1)]$
$= \frac{1}{3} [\mathrm{f}(n) - \mathrm{f}(0)]$
$= \frac{1}{3} \left[ \left(1 - \frac{3}{(n+2)!}\right) - \left(1 - \frac{3}{2}\right) \right] = \frac{1}{2} - \frac{1}{(n+2)!}$

2. Use an index substitution to relate the sum in part (ii) to the sum in part (i).

We need to evaluate $\sum_{r=3}^{n} \frac{r-1}{r!}$.
Let $k = r - 2$. The limits change accordingly:
When $r = 3$, $k = 1$.
When $r = n$, $k = n - 2$.
Substituting this into the summation:
$\sum_{r=3}^{n} \frac{r-1}{r!} = \sum_{k=1}^{n-2} \frac{(k+2)-1}{(k+2)!}$
$= \sum_{k=1}^{n-2} \frac{k+1}{(k+2)!}$

3. Apply the result from part (i) to deduce the final value.

Using the established result $\sum_{r=1}^{N} \frac{r+1}{(r+2)!} = \frac{1}{2} - \frac{1}{(N+2)!}$ with the new upper limit $N = n - 2$:
$\sum_{k=1}^{n-2} \frac{k+1}{(k+2)!} = \frac{1}{2} - \frac{1}{((n-2)+2)!}$
$= \frac{1}{2} - \frac{1}{n!}$