(a) We start with \(\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} (r^2 + 2r)\).

Using the formulae, \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\).

Thus, \(\sum_{r=1}^{n} r(r+2) = \frac{1}{6}n(n+1)(2n+1) + n(n+1)\).

Simplifying gives \(\frac{1}{6}n(n+1)(2n+7)\).

(b) Express \(\frac{1}{r(r+2)}\) in partial fractions: \(\frac{1}{r(r+2)} = \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right)\).

Then, \(\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{n} - \frac{1}{n+2} \right)\).

This simplifies to \(\frac{1}{2} \left( \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the terms \(\frac{1}{n+1}\) and \(\frac{1}{n+2}\) approach zero.

Thus, the sum converges to \(\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}\).