(a) We start with \(\sum_{r=1}^{n} r(r+1)(r+2) = \sum_{r=1}^{n} (r^3 + 3r^2 + 2r)\).

Using standard summation formulae:

\(\sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2\)

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

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

Substituting these into the expression:

\(\frac{1}{4}n(n+1)^2 + \frac{1}{2}n(n+1)(2n+1) + n(n+1)\)

Factorising gives \(\frac{1}{4}n(n+1)(n+2)(n+3)\).

(b) Express \(\frac{1}{r(r+1)(r+2)}\) in partial fractions:

\(\frac{1}{r(r+1)(r+2)} = \frac{1}{2r} - \frac{1}{r+1} + \frac{1}{2(r+2)}\)

Using the method of differences:

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

This telescopes to:

\(\frac{1}{2}f(1) - f(2) + \frac{1}{2}f(3) + \frac{1}{2}f(2) - f(3) + \frac{1}{2}f(4) + \ldots + \frac{1}{2}f(n) - f(n+1) + \frac{1}{2}f(n+2)\)

Resulting in \(\frac{1}{4} \left( \frac{1}{n+1} + \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the sum becomes \(\frac{1}{4}\).