(a) Start by expressing \(\frac{1}{(kr+1)(kr-k+1)}\) using partial fractions:

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

Thus, the sum becomes:

\(\sum_{r=1}^{n} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( 1 - \frac{1}{k+1} + \frac{1}{k+1} - \frac{1}{2k+1} + \ldots + \frac{1}{k(n-1)+1} - \frac{1}{kn+1} \right)\)

This simplifies to:

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

(b) For the infinite sum, as \(n \to \infty\), the term \(\frac{1}{kn+1} \to 0\), so:

\(\sum_{r=1}^{\infty} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\)

(c) Using the result from part (a), the sum from \(n\) to \(n^2\) is:

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

Using the formula from part (a):

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

This simplifies to:

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