It is given that $\text{f}(n) = \frac{n}{5^{n-1}}$ where $n$ is a positive integer.

1. Evaluate and simplify $\text{f}(r) - \text{f}(r+1)$ to show its relationship to the summand.

$\text{f}(r) - \text{f}(r+1) = \frac{r}{5^{r-1}} - \frac{r+1}{5^r}$
$= \frac{5r}{5^r} - \frac{r+1}{5^r}$
$= \frac{4r - 1}{5^r}$

2. Express the summation using the method of differences and cancel the intermediate terms.

$\sum_{r=2}^{n} \frac{4r - 1}{5^r} = \sum_{r=2}^{n} \big( \text{f}(r) - \text{f}(r+1) \big)$
$= \big( \text{f}(2) - \text{f}(3) \big) + \big( \text{f}(3) - \text{f}(4) \big) + \dots + \big( \text{f}(n-1) - \text{f}(n) \big) + \big( \text{f}(n) - \text{f}(n+1) \big)$
$= \text{f}(2) - \text{f}(n+1)$

3. Substitute the specific values into the function $\text{f}$ to obtain the final simplified expression.

$\text{f}(2) = \frac{2}{5^{2-1}} = \frac{2}{5}$
$\text{f}(n+1) = \frac{n+1}{5^{(n+1)-1}} = \frac{n+1}{5^n}$
$\sum_{r=2}^{n} \frac{4r - 1}{5^r} = \frac{2}{5} - \frac{n+1}{5^n}$

1. State the result from part (a), finding the expression for the sum by the method of differences.

First, evaluate the expression established in part (a):
$\text{f}(r) - \text{f}(r+1) = \frac{r}{5^{r-1}} - \frac{r+1}{5^r} = \frac{5r - (r+1)}{5^r} = \frac{4r - 1}{5^r}$
$\sum_{r=2}^{n} \frac{4r-1}{5^r} = \text{f}(2) - \text{f}(n+1) = \frac{2}{5} - \frac{n+1}{5^n}$

2. Apply an index substitution to match the denominator in the target sum with part (a).

Let $k = r+1$. When $r=1$, $k=2$; when $r=n$, $k=n+1$.
$\sum_{r=1}^{n} \frac{4r+6}{5^{r+1}} = \sum_{k=2}^{n+1} \frac{4(k-1)+6}{5^k} = \sum_{k=2}^{n+1} \frac{4k+2}{5^k}$

3. Split the modified sum into a term resembling part (a) and a separate term.

$\sum_{k=2}^{n+1} \frac{4k+2}{5^k} = \sum_{k=2}^{n+1} \frac{4k-1+3}{5^k} = \sum_{k=2}^{n+1} \frac{4k-1}{5^k} + \sum_{k=2}^{n+1} \frac{3}{5^k}$

4. Evaluate the first summation using the result from part (a).

Replacing $n$ with $n+1$ in the result derived in part (a):
$\sum_{k=2}^{n+1} \frac{4k-1}{5^k} = \text{f}(2) - \text{f}(n+2) = \frac{2}{5} - \frac{n+2}{5^{n+1}}$

5. Evaluate the second summation as a finite geometric series.

The series $\sum_{k=2}^{n+1} \frac{3}{5^k}$ is a geometric progression with first term $a = \frac{3}{25}$, common ratio $r = \frac{1}{5}$, and $n$ terms.
$S_n = \frac{\frac{3}{25} \left(1 - \left(\frac{1}{5}\right)^n\right)}{1 - \frac{1}{5}} = \frac{\frac{3}{25}}{\frac{4}{5}} \left(1 - \frac{1}{5^n}\right) = \frac{3}{20} \left(1 - \frac{1}{5^n}\right)$

6. Combine the parts and simplify algebraically to obtain a single expression.

Summing the evaluations together:
$\text{Total sum} = \frac{2}{5} - \frac{n+2}{5^{n+1}} + \frac{3}{20} - \frac{3}{20 \cdot 5^n}$
Notice that $\frac{3}{20 \cdot 5^n} = \frac{3}{4 \cdot 5 \cdot 5^n} = \frac{3}{4 \cdot 5^{n+1}}$. Gather constant terms and terms with $5^{n+1}$ in the denominator:
$= \left(\frac{2}{5} + \frac{3}{20}\right) - \left( \frac{4(n+2)}{4 \cdot 5^{n+1}} + \frac{3}{4 \cdot 5^{n+1}} \right)$
$= \frac{11}{20} - \frac{4n+8+3}{4 \cdot 5^{n+1}}$
$= \frac{11}{20} - \frac{4n+11}{4 \cdot 5^{n+1}}$