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

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

Now, apply the method of differences:

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

This simplifies to:

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

(b) For the infinite series, the terms after the first cancel out, leaving:

\(\frac{1}{k}\)

(c) Use the same method of differences for the range \(r=n\) to \(r=n^2\):

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

This results in:

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