(a) Start by expressing \(\frac{5k}{(5r+k)(5r+5+k)}\) using partial fractions:

\(\frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5r+k} - \frac{1}{5r+5+k} \right)\)

Then, the sum becomes:

\(\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5+k} + \frac{1}{10+k} + \frac{1}{15+k} + \ldots + \frac{1}{5n+k} - \frac{1}{5+k} - \frac{1}{10+k} - \ldots - \frac{1}{5n+5+k} \right)\)

By the method of differences, this simplifies to:

\(k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

(b) Given \(\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}\), we have:

\(k \left( \frac{1}{5+k} \right) = \frac{1}{3}\)

Solving for \(k\):

\(\frac{k}{5+k} = \frac{1}{3}\)

\(3k = 5 + k\)

\(2k = 5\)

\(k = \frac{5}{2}\)

(c) Using the result from (a), find:

\(\sum_{r=n}^{n^2} \frac{5k}{(5r+k)(5r+5+k)} = \sum_{r=1}^{n^2} \frac{5k}{(5r+k)(5r+5+k)} - \sum_{r=1}^{n-1} \frac{5k}{(5r+k)(5r+5+k)}\)

\(= k \left( \frac{1}{5+k} - \frac{1}{5n^2+5+k} \right) - k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

\(= k \left( \frac{1}{5n+5+k} - \frac{1}{5n^2+5+k} \right)\)

Substitute \(k = \frac{5}{2}\):

\(= \frac{5}{2} \left( \frac{1}{5n+5+\frac{5}{2}} - \frac{1}{5n^2+5+\frac{5}{2}} \right)\)

\(= \frac{1}{2n+1} - \frac{1}{2n^2+3}\)